Thank you Andy. I am a British maths teacher (well, I was until yesterday, when I retired) and watch lots of math solutions - I have given up watching many channels similar to this, as they take so long to get to the point. I like Numberphile and similar channels but this is the only geometry problem one I follow now as Andy zips through the explanation very quickly and clearly.
Same her without the being or having been a math teacher part (well, except for my girlfriend who needs math for her studying medicine and has a math phobia).
I solved it a different way: 1. Draw a radius to the point the circle touches the square bottom left of it. 2. Make this the hypotenuse of a 45-45-90 triangle within the circle. 3. If we call the radius x, both legs of the triangle must be x/√2 (because of the property of a 45-45-90 triangle where the hypotenuse = leg√2). 4. Draw another radius directly right, to the point the circle touches the outer box. 5. If you look at the horizontal leg of your triangle and the radius drawn in step four, you can see that the two distances add up to create the distance labelled underneath to be 2. 6. With this, you can create the equation: x/√2 + x = 2 7. solve for x and you get x = 4/(2 + √2), or x = 1.172
I caluculated it differently. We know that 2 lines originating from the same point, touching the same cirlce have the same length. If we draw a diagonal from the bottom right to touch the circle, we know that its 2√2. That means the verticle line that touches the circle is also 2√2. Then we just subtract that from the full length of the side giving us 4-2√2 which comes out to the correct number
There's another way for solving this. If you draw a line between the top-left and bottom-right vertex of the square you get a right-angled triangle with the circle being its inscribed circle. Let P be the perimeter of the triangle and A its Area. The radius of the inscribed Circle (inside a right-angled triangle) is equal to 2A/P
I did something similar, but instead of the inscribed circle property, I used the top letf corner going right until it tangents the circle, and then a diagonal line from the same corner to the center of the big square, creating another tangent line that must equal each other in length, since they have the same starting point. From there, 4 = 2sqrt2 [which is the tangent segment length] + r So R = 4-2sqrt2 =~1.17
I only discovered your channel a week ago, you've ignited a love for maths I never knew I had. Thanks for the enjoyable videos. all the best, from Australia :)
It is easier to draw a line from the upper left vertex to the central point 2,2 which is tangent to the circle. Knowing that two tangents to a circle from a common point are equal, the upper horizontal tangent will be equal to the diagonal 2sqrt(2). Thus, 2sqrt(2)+r=4. Then r=4-2sqrt(2).
r = 2√2/(1+√2) = 4 - 2√2 (rationalizing the denominator) ≈ 1.1716 I solved it somewhat convolutedly I drew the square in the top left with radius r at the tangent points, then extended the vertical and horizontal lines of that square to make a + shape. I also extended the lines of the bottom-left square to make a + shape I end up with two small triangles of sides (2-r):(2-r):r, with radius r as the common hypotenuse then I used a²+b²=c² on one of the triangles to solve for r. (2-r)² + (2-r)² = r² you can also compare it to a similar 1:1:√2 triangle: (2-r)/r = 1/√2
I solved this an (i think) easier way. From the center of the circle, you can draw a line straight to one of the edges of the outer square, then another line at 45° to the corner of the inner square. Since we know that the height/width of these lines as one shape is 2, you can use the equation r + rsin(45°) = 2 to solve for r and get the same answer
I did it similarly to this, but instead of using sin(45°), I just plugged the value of sqrt(2)/2 because the pithagorean theorem gave me that number. Then rationalized to: 4 - 2sqrt(2), which gave me the answer
Ok, now I think I understood your solution. You can also draw a line from the center of the circle to the common angle of the squares and another one parallel with a side of little square and toward a common line of two squares. You get again 2 Δ that are proportional. So you get (4-r)/2 = (sqr8+r)/sqr8 .
It's been decades since I finished high school, and I'm still in love with these kind of problems. Better than real world problems. Keep up the great work.
Here’s my thinking before watching yours. Set origin at upper right of 2x2 square. Center position can either be represented (in both horizontal and vertical) as 2-r or r/√2, so these are equal statements. Solving 2-r=r/√2 gives r=4-2√2
I'm a big believer in rationalizing the denominator. Sometimes it makes it better. When you start factoring cubics and bigger, it's a good skill to have.
I like the notion of a general solution with such problems. Looking at the sketch, the half of the square's diagonal is s·√2/2, which is equal to r + r√2, so one gets s·√2/2 = r(1 + √2) → r = s·√2/(2·(1 + √2)). Then we can rationalize this to get r = (s/2)·√2/(1 + √2)·[(1 - √2)/(1 - √2)], hence r = (s/2)·(√2 - 2)/(1-2) → r = s·(2-√2)/2. Et voilà!
Pretty simple task. We have two squares: one with side 2, one with side r. Their diagonals are square root of 2 times 2 and square root of 2 times r. Add up these diagonals, add another r and you get the diagonal of the biggest square which is square root of 2 times 4. So we have: 4•2^1/2 = 2•2^1/2 + r•2^1/2 + r r(1 + 2^1/2) = 4•2^1/2 - 2•2^1/2 r = (2•2^1/2)/(1 + 2^1/2) r = 2,818 / 2,414 = 1,167
1. Consider lower left corner to be (0,0). Draw line from (0,0) to (4,4). 2. Mark circle center as (x,y). Draw radius R from (2,2) to (x,y). 3. Draw horizontal line from (2,2) to (x,2). Label as A' 4. Draw vertical line from (x,2) to (x,y). Label as A 5. Draw horizontal line from (x,y) to the right to circle square tangency point. Label as R. 6. Angle at (2,2) = 45. Solve for A. A = R*cos(45) 7. A + R = 2. R*cos(45) + R = 2. 8. R(cos(45) +1) = 2. 9. R = 2/cos(45) + 1 = 1.1715
After calculating 2√2 as half of the diameter, we should remember that the two tangents drawn from a point to a circle are equal. So the tangent line drawn from the top left corner of the square to the circle is also 2√2. Finally, subtract 4 - 2√2, and you'll get the same answer.
Each time you solve one of these, I almost always don't figure it out before you do... And that both irritates me, and humbles me at the same time 😜.. You rock... I'm showing this to my students as well.
I totally trig'd it knowing it is just a 45 deg angle, thus basically immediately got r(1 + sqrt(1/2)) = 2 and thus simplified to r = 4 - 2 sqrt(2). Obviously same result as you got, just with the usual trick of expanding the fraction by the conjugate of the denominator.
Faster way: The point of the circle that touches the small square is 2 units from the right, and we have a radius going horizontally plus a radius at a 45deg diagonal to make that 2-unit horizontal, meaning 1r + r/sqrt(2) = 2 => r = 2/(1+1/sqrt(2)) = 1.17157
connect diagonal from top left to bottom right, it will be tangent to the circle and form a triangle then use formula rp=S where p is a semiperimeter, S is an area of a triangle and r is radius you can find the area of triangle by using heron's formula and a semiperimeter by summing up three sides and dividing them by 2.
I’m with you…how exciting! When I help my teenager with math problems, I’ve started saying “how exciting” when we get to a solution. Not sure it’s working but I keep trying!
Please always rationalize the denominator. You'll get rid of the fraction and you'll only have one root in your answer: 4 - 2sqrt(2) is so much nicer and simpler.
Just draw the diagonal of the big square from top left to bottom right. You get an isosceles triangle and the circle will be it's incircle. Then it's easy to calculate the radius.
This seemed like one of those questions on a math test I would skip because it took way to many steps and not enough time to deal with it when you have 3 other questions that are just as long but are easier to solve.
i took a different approach as follows: First, i took the point of contact between the circle and the top right corner of the inner square, and make it the origin point of a X,Y coordinate system. Then the general equation for that circle in that X,Y coordinate system would be (X-a)^2 + (Y-b)^2=r^2 where "a" and "b" are the horizontal and vertical distances from the center of the circle to the origin point. And because the circle is bisected by the diagonal of the big square, we can asume "a" and "b" are equal, and we call this value c. Thus, the equation now comes to (X-c)^2+(Y-c)^2=r^2 Now, we know when x=0 then y=0 wich leads to r=c√2 (1) Then, we also know that for x=c then y=2 wich leads to c=2-r (2) Then replacing (2) in (1) we get r=(2√2)/(1+V2) ---> r=1,17
i solved it differently, i made a right angled triangle using from the center of the circle to the top right diagonal of the square. my hypotenuse was r and my side lengths were 2-r. then afterwards i quad formula the quadratic and got 4-2root2 which gives the right answer
I did it a bit different, we can ignore the big square and just consider the right top quarter, and since the circle is limited on one side by the edges of that square and on the other side by the corner of that square, then we can draw 2 more circles: 1. draw a circle that goes thru all corners of the square, find the diameter, here ofc the diameter of the circle is the diagonal of the square, so same as in the video, using pytagora -> 2 root(2) 2. draw another circle, that fits entirely inside the square, a.k.a. inscribed in the square -> here we can see the diameter is the length of the circle = 2 -> and we can see the difference between the diameter of these 2 new circles is 'x' on both sides, so big circle diameter = small circle diameter + 2x => 2 root(2) = 2 + 2x => x = root(2) - 1 -> we can also see that the difference between the diameter of the new smaller circle and the original circle is just 'x', so original circle diameter = small circle diameter + x => d = 2 + x => d = 2 + root(2) - 1 => d = 1 + root(2) -> now we can find the radius by dividing the diameter with 2 => r = (1 + root(2)) / 2 = 1.2071
I went about this differently and got the same result. Draw a diagonal line from the upper left corner(call it A) to the lower right corner of the large square. Find the point of the circle that touches this diagonal (call it B). The length from the upper corner to point A is the diagonal of the square, which is 2 x sqrt(2). Find the point where the circle meets the upper side of the square (call it C). AB = AC. r = 4 - AC = 4 - 2 x sqrt(2) = 1.17.157
Why are you using Pythagoras for all those diagonals ? It would be faster to use the formula d = s * √(2) where d is the diagonal of the square et s its side.
Or more precisely, 4 - sqrt(8). Co-ordinate type geometry is an easy way to see this. Centre of the circle is (x,y). We have two radii from (x,y) to (4,y) and (x,4). The radius is therefore 4-x (or 4-y), so we see x = y, as we'd expect from the symmetry of the construction. Now we have another radius from (2,2) to (x,y). So by Pythagoras we have 2(x-2)^2 = radius^2 = (4-x)^2. Solve this to get x = sqrt(8) (clearly x>0 by construction) and therefore the radius is 4-sqrt(8).
More cleanly, 2*sqrt(2)/(1+sqrt(2)) = 4-2*sqrt(2). So... that means the diagonal between that bottom right corner and the center point of the circle is exactly 4. In case anyone was interested in a more intuitive way of looking at the problem.
I guess I did this one the hard way. First calculate the diagonal of the small square, as shown. Then extend that line to the center of the circle to create a right triangle with hypotenuse 2√2 + r. Each leg is 4 - r. The Pythagorean Theorem means that (4 - r)^2 + (4 - r)^2 = (2√2 + r)^2. Multiply it out and you get a knotty quadratic: r^2 - (16 + 4√2)r + 24 = 0. This actually is amenable to solving with the quadratic formula. You get a root within a root, but that can be handled. Finally, r = [(16 + 4√2) +/- (8 + 8√2)]/2. The larger root is way too big for the given parameters. But the smaller root, 4 - 2√2, does equal 1.17157.
The radius of a circle inscribed in a triangle is two times the area of the triangle divided by the perimeter of the triangle. The area of the triangle is 8. The perimeter of the triangle is 4 + 4 + 4xsqrt(2). The radius of the circle is 16/(8+4xsqrt(2))=1.17157.
I loved multi step problems like these in school. It felt like going on a bunch of side quests in a game. And it actually took some time to solve, which meant less time bored in class.
lol I too a completely different approach (didn’t have pen/paper so I needed an easier way): basically, assume the center of the circle is (x,y). The center of the circle must lie on the main ascending diagonal of the square for symmetry reasons, so x=y. The lower square touches the circle at (2,2). Using this, the distance from (2,2) to (x,x) is sqrt(2)*(x-2)… But going from the center of the circle to the top of the large square is also a radius (4-x)… you can do the same horizontally. Thus, we get sqrt(2)*(x-2)=4-x. Notice that this is a simple linear equation in one variable… solve for x we immediately get: [sqrt(2)+1]*x = 4+sqrt(2)*2 x = 2[sqrt(2)+2]/[sqrt(2)+1] = 2*sqrt(2) … therefore, radius = 4 - 2*sqrt(2) 🥳 ✨as a bonus this mean’s the CZcamsrs expression 2*sqrt(2)/[1+sqrt(2)] simplifies beautifully to 4 - 2*sqrt(2)
I’m going to start doing these problems with my kids because I think if they saw how you legitimately enjoy solving these problems they could learn to do the same, and I actually enjoy math when it is approached like this. I honestly wish you all the happiness in the world brother. You’re making a difference.
I use BPT theorem to find radius of the circle by making two equal triangel with common hypotenious which is equal to the diameter of the square , and let the base of these two triangle by making one horizontal and vertical line from the side which is touching with the circle which makes the two triangle who follow the BPT rule because the length "r" both horizontal and vertical and "2units" both horzontal and vertical are prallei t each other . Now i put the values directly and indirectly and apply the BPT to get radius thanks for giving such cute question ❤🔥❤🔥 edit= my approx anser is |1.18| because i use"root2" as 1.14
That is one long way to do it. Or draw a line from one corner to the other where that forms a triangle with the circle inside it, and the circle, that line (hypotenuse H), and the top right corner of the little square all touch. Then the formula for the max radius of a circle that can fit in a triangle is r=(P + B - H)/2 P and B (perpendicular and base) r= ( 4 + 4 - 4sqrt(2) ) / 2 r = ( 8 - 4sqrt(2) ) / 2 r = 4 - 2sqrt(2)
Solved it a different way: 1. Rotated the whole thing 180 degrees 2. Now it looks like the co-ordinate system 3. One side of square as X-axis and the other as Y-axis 4. We can clearly see that the circle is passing through the point (2,2) 5. Now it's just a simple question that is taught to us "A circle is passing through a point (2,2) and touches both the axes, find the radius" 6. If we take x^2 + y^2 + 2gx + 2fy + c as the general equation of the circle, the radius become sqrt(g^2+f^2-c) 7. The circle touches both the axes, so g^2 = f^2 = c, Hence the radius = √(g^2+g^2-g^2) = g = f = √(c) 8. Put that in the equation, x^2 + y^2 + 2√cx + 2√cy + c = 0 9. Substitute the point (2,2) in the equation and take squaring on both sides. You get a quadratic and put that in calculator and you get c. Now we need √c so just sqrt the number you get and that's the answer.
Its not the fact he solves these things - I dont think that I’ve ever seen him do maths that I didnt know - but its the ingenuity with which he solves it. Its one thing knowing how to get through the math, and another thing entirely on seeing a bunch of shapes and knowing what math to use to get you there.
I just put a diagonal in the square from the top left to the bottom right made the sides equal to n+r and the diagonal to 2n since we have 90° we know that the diagonal is equal to √(2×4²) meaning n is half of that than just 4-n and you have r. If you continue with this you can get that r is about equal to 0,29289[((2-√(2))÷2), 1-√0.5]×a side of the big square for all values if you have different side lengths it would obviously change, even tho you would still be able to generalize that.
My answer: 4-2root2 My working: The vertical distance from the corner of the little square to the roof of the big square is 2 The vertical distance from where the circle touches the corner of the little square to the roof of the big square is radius + (radius/root(2)) since the angle is 45 degrees Then just do the algebra for 2 = r + r/root2
An alternative form of the solution is 4-√8, now solve the 3D equivalent of the excercise (A half size cube inside a cube and a spehere touching its corner and the faces of the bigger cube, what is its radius?)
For any triangle S=p*r, where p - half perimeter, r - radius of inscribed circle. In our case, S = 0.5*4*4 = 8 (half the square); p = 0.5*(4+4+4*sqrt2). Thus, r = 8/(4+2*sqrt2) = 4/(2+sqrt2) = 2*(2-sqrt2) =~ 1.1715
My approach to this problem is different we don't need the square, divide the squared into 2 parts diagonally and this should give you a circle inscribed in a triangle then therefore we can use the formula r=(a+b-c)/2 a=b and c=4root2 r=(4+4-4root2)/2 or 1.1715...
The diameter of the circle is larger than 2. Assuming the circle is tangent to the outer box in both parts of the corner that diagonal line would never intersect the center point of the circle. so your answer is close but not correct
I have a faster way, but it requires knowledge of the quadratic formula or completing the square. Andy’s solution is more convoluted, but utilizes simpler techniques that more people will understand 👍🏼
Exellent solution! I am trying to make understanding of this, and 1:30 you said you can sqaure root both sides of the equation. Unfortunately I don't think this works as 3^2 * 4^2 = 5^2, however 3 * 4 does not equal five. I would be gratefull if you could shine some insight on this.
Draw a Tangent to the circle from bottom right corner of Larger Square(where tangent touches the circle at centre of the larger square) and this length is half the diagonal of larger square = 2√2. Now tangents drawn from a point outside to the circle are equal in length(one along the diagonal and one along side from the bottom corner of larger square) hence 2√2 + r = 4 r = 4 - 2√2 OR Draw the diagonal of the larger square (connecting bottom right corner and top left corner vertices). Now we have a Right Triangle where the sides are 4, 4, 4√2 and circle is the incircle of this triangle whose radius is (4+4-4√2)/2 = 4-2√2 PS :: Inradius of a Right Triangle, whose sides are a,b,c with c being hypotenuse is (a+b-c)/2
Last advice to use calculator would lower the score when I was in school. My teachers always emphasized to use exact computation, in this case it means that fraction should be multiplied (upper and lower part) by (sqrt(2)-1). Then upper part becomes 4-2*sqrt(2) and lower part becomes 1 (because (a+b)*(a-b) = a^2-b^2, it his case a=sqrt(2), b=1 so a^2-b^2=2-1=1). Final value should be 4-2*sqrt(2).
If the goal was to solve this problem using the most complicated method possible it was a success. It's a lot less calculation intensive to solve this problem (starting from the drawing with the 3 radii) by simply adding a diagonal line from the top left corner of the square to the center of the square which is a point of tangency with the circle. Because we can see that this diagonal line is the hypotenuse of an isosceles right triangle with side lengths of 2, we know that this hypotenuse length will equal 2√2 and therefore the length of the square's top side left section (from the left corner to the point of tangency at the top of the circle) is also equal to 2√2. The drawing has already established that the square's top side right section (from the point of tangency at the top of the circle to the right corner) is equal to the length of the radius. The length of the radius equivalent section of the top side of the square (r) can be calculated from the difference between the total length of the top side of the square (4) and the length of the left section of the top side of the square (2√2). r = 4 - 2√2 r ≈ 1.17
My solution: - The large square has a side of 4. Take 2 from the small square. We have now 2. - From the circle we have r+r*cos(45°) which is equal to 2 Then r+r*cos(45°)=2 r(1+sqr(2)/2)=2 -> r≈1.1716
The top right point in the square is the director circle to the given circle now using coordinate geometry results we know distance of such point from center of circle is always r√2 Director circle : locus of points from which tangents drawn to a curve are mutually perpendicular Where r is the radius of given
Hi sir, I highly admire the way you solve math problems it emphasizes my love toward solving complex problems. Thereby can you make a pdf (free for veiwers) based on some questions like these amd other of your math problems for guys like me to solve.
Logic flaw here, if you rotate side 'c' down onto side side it 'must be longer' than side 'a', if it were actually 2 units the angle of 'ca' would be zero or side 'a'. Simple proof on paper, create the box, draw the diagonal 'c', get a protactor (Lead pencil) spead the legs the distance of 'a' and rotate the protractor to the verticle, 'c' should be bisected.
You should have rationalized the denominator by multiplying top and bottom by (1 - sqrt(2)). The denominator simplifies to 1, which eliminaties the fraction completely, and the numerator becomes 4 - 2(sqrt(2)). Evaluating that gives the same answer: 1.17157...etc. Also, much simpler way to do this: Extend the horizontal top of the lower left square to the right-hand vertical side, dividing it into 2 segments, each length 2 (just like the left side). The top part of the top segment is r, as you've shown. The bottom part of that top segment is the side of a square whose diagonal is r. That's r/sqrt(2). So now we have r + r/sqrt(2) = 2. Solving this equation gives the same result.
I think he should have taken a different variable name apart form c for finding the hypotenuse using the circle with radius r as he had already used variable c for the square of sides 2
I suck at math but unless I am mistaken the radius of that circle has to be half the distance from the connecting corner of the lower right square to the square to the opposite corner of the upper right square and that is simply the square root of 2 squared + 2 squared = 2.828 no?
I find the abswer. first make a triangle by connecting diagonal area of triangle is 4 x 2 = 8 the perimeter of triangle is 4√2 + 8, so S = 2√2 + 4 R = 8 : (2√2 + 4) = 4/ ( √2+2)
He wasn’t as excited for this 😢😢🤧
Maybe because it was too easy for him and was not challenging? 😅😂😂😅
@@bebektoxic2136sad since it took me like 10 minutes to solve it
That's not the point, doesn't matter if he is excited or not.
@@arpieravidas7731 it is the point, I’m heartbroken
I hope Mr Andy Math makes a full recovery for the next one 😔💔
Thank you Andy. I am a British maths teacher (well, I was until yesterday, when I retired) and watch lots of math solutions - I have given up watching many channels similar to this, as they take so long to get to the point. I like Numberphile and similar channels but this is the only geometry problem one I follow now as Andy zips through the explanation very quickly and clearly.
isn't it the ultimate proof that math teachers retire from school, but never from math? absolute legend.
Congratulations and happy retirement! I'm 4 years in to maths teaching. Unsure if I'll carry on!
Amen he is alive!
Same her without the being or having been a math teacher part (well, except for my girlfriend who needs math for her studying medicine and has a math phobia).
I solved it a different way:
1. Draw a radius to the point the circle touches the square bottom left of it.
2. Make this the hypotenuse of a 45-45-90 triangle within the circle.
3. If we call the radius x, both legs of the triangle must be x/√2 (because of the property of a 45-45-90 triangle where the hypotenuse = leg√2).
4. Draw another radius directly right, to the point the circle touches the outer box.
5. If you look at the horizontal leg of your triangle and the radius drawn in step four, you can see that the two distances add up to create the distance labelled underneath to be 2.
6. With this, you can create the equation: x/√2 + x = 2
7. solve for x and you get x = 4/(2 + √2), or x = 1.172
Step 6: If you draw the "radius square" at the top right, you can ultimately see that x + x/√2 is half of the outer square's diagonal 😀
That's the method I used when I saw the thumbnail. Glad I'm not alone.
1. draw it in scale and measure the radius
I wish I could get high off potenuse 🤓
Here's how to solve x:
x/√2 + x = 2
x*√2 + x*√2*√2 = 2*√2*√2 multiply both sides by √2*√2
x*√2 + x*2 = 4 simplify (√2*√2 = 2)
x*(√2 + 2) = 4 factorize
x = 4/(√2 + 2) solve x
I caluculated it differently. We know that 2 lines originating from the same point, touching the same cirlce have the same length. If we draw a diagonal from the bottom right to touch the circle, we know that its 2√2. That means the verticle line that touches the circle is also 2√2. Then we just subtract that from the full length of the side giving us 4-2√2 which comes out to the correct number
Very clever
There's another way for solving this.
If you draw a line between the top-left and bottom-right vertex of the square you get a right-angled triangle with the circle being its inscribed circle.
Let P be the perimeter of the triangle and A its Area.
The radius of the inscribed Circle (inside a right-angled triangle) is equal to 2A/P
it's interesting how more knowledge leads to more elegant solutions
I did something similar, but instead of the inscribed circle property, I used the top letf corner going right until it tangents the circle, and then a diagonal line from the same corner to the center of the big square, creating another tangent line that must equal each other in length, since they have the same starting point. From there, 4 = 2sqrt2 [which is the tangent segment length] + r
So R = 4-2sqrt2 =~1.17
This is how I did it too and to me feels more direct
Everything is temporary but "how exciting" is permanent 😂😂
The radius of a circle inscribed in a right triangle is (a+b-c)/2 = (4+4-4√2)/2 = 4-2√2.
I only discovered your channel a week ago, you've ignited a love for maths I never knew I had. Thanks for the enjoyable videos.
all the best, from Australia :)
my fellow upside down mate.
Exactly the same for me, I’m actually enjoying maths, which is something i never thought I’d say. Greetings from Germany!
We making it out of math class with this one 🔥🔥🔥💯💯💯
Facts fr
u got the whole squad laughing
seeing a math problem from you on my feed always makes me happy to try it
It is easier to draw a line from the upper left vertex to the central point 2,2 which is tangent to the circle. Knowing that two tangents to a circle from a common point are equal, the upper horizontal tangent will be equal to the diagonal 2sqrt(2). Thus, 2sqrt(2)+r=4. Then r=4-2sqrt(2).
r = 2√2/(1+√2)
= 4 - 2√2 (rationalizing the denominator)
≈ 1.1716
I solved it somewhat convolutedly
I drew the square in the top left with radius r at the tangent points, then extended the vertical and horizontal lines of that square to make a + shape. I also extended the lines of the bottom-left square to make a + shape
I end up with two small triangles of sides (2-r):(2-r):r, with radius r as the common hypotenuse
then I used a²+b²=c² on one of the triangles to solve for r. (2-r)² + (2-r)² = r²
you can also compare it to a similar 1:1:√2 triangle: (2-r)/r = 1/√2
I solved this an (i think) easier way. From the center of the circle, you can draw a line straight to one of the edges of the outer square, then another line at 45° to the corner of the inner square. Since we know that the height/width of these lines as one shape is 2, you can use the equation r + rsin(45°) = 2 to solve for r and get the same answer
I like that one, it makes sense to me though it didn't occur to me. I really need to brush up on my trig, haven't done it since the 90's
I did it similarly to this, but instead of using sin(45°), I just plugged the value of sqrt(2)/2 because the pithagorean theorem gave me that number. Then rationalized to:
4 - 2sqrt(2), which gave me the answer
By your method we can solve in other two ways:
1. The center of that circle is the intersections of the bisectors of your triangle. A bisector from a
Nevermind, I didn't read carefully your thing. I thought you refered to a triangle with its hypotenuse as the other diameter of the big square.
Ok, now I think I understood your solution. You can also draw a line from the center of the circle to the common angle of the squares and another one parallel with a side of little square and toward a common line of two squares. You get again 2 Δ that are proportional. So you get (4-r)/2 = (sqr8+r)/sqr8 .
Fun, indeed. Thanks. 80 y/o here who used to have fun doing math. I'll be watching you channel regularly.
Multiply the top and bottom by (1-√2)(-1).
The denominator becomes 1, and the numerator becomes 4-2√2.
Change from 1+√2
To (√2)+1 because it's the same
It's been decades since I finished high school, and I'm still in love with these kind of problems. Better than real world problems. Keep up the great work.
Here’s my thinking before watching yours.
Set origin at upper right of 2x2 square.
Center position can either be represented (in both horizontal and vertical) as 2-r or r/√2, so these are equal statements.
Solving 2-r=r/√2 gives r=4-2√2
I'm a big believer in rationalizing the denominator. Sometimes it makes it better. When you start factoring cubics and bigger, it's a good skill to have.
I like the notion of a general solution with such problems. Looking at the sketch, the half of the square's diagonal is s·√2/2, which is equal to r + r√2, so one gets
s·√2/2 = r(1 + √2) → r = s·√2/(2·(1 + √2)). Then we can rationalize this to get
r = (s/2)·√2/(1 + √2)·[(1 - √2)/(1 - √2)], hence
r = (s/2)·(√2 - 2)/(1-2) → r = s·(2-√2)/2. Et voilà!
Your how exciting was rushed for this one! Thanks for another fun problem.
Pretty simple task. We have two squares: one with side 2, one with side r. Their diagonals are square root of 2 times 2 and square root of 2 times r. Add up these diagonals, add another r and you get the diagonal of the biggest square which is square root of 2 times 4. So we have:
4•2^1/2 = 2•2^1/2 + r•2^1/2 + r
r(1 + 2^1/2) = 4•2^1/2 - 2•2^1/2
r = (2•2^1/2)/(1 + 2^1/2)
r = 2,818 / 2,414 = 1,167
1. Consider lower left corner to be (0,0). Draw line from (0,0) to (4,4).
2. Mark circle center as (x,y). Draw radius R from (2,2) to (x,y).
3. Draw horizontal line from (2,2) to (x,2). Label as A'
4. Draw vertical line from (x,2) to (x,y). Label as A
5. Draw horizontal line from (x,y) to the right to circle square tangency point. Label as R.
6. Angle at (2,2) = 45. Solve for A. A = R*cos(45)
7. A + R = 2. R*cos(45) + R = 2.
8. R(cos(45) +1) = 2.
9. R = 2/cos(45) + 1 = 1.1715
You lost me at “fun geometry challenge”
This is actually a Trigonometry challenge
That was the most genuine "how exciting" I've heard you proclaim. Hyped me up to make a comment.
One can also solve through equation of circle by taking bottom left as origin, centre lies on y=x and y=4 and x=4 are tgts to circle
After calculating 2√2 as half of the diameter, we should remember that the two tangents drawn from a point to a circle are equal. So the tangent line drawn from the top left corner of the square to the circle is also 2√2. Finally, subtract 4 - 2√2, and you'll get the same answer.
Each time you solve one of these, I almost always don't figure it out before you do... And that both irritates me, and humbles me at the same time 😜..
You rock... I'm showing this to my students as well.
I totally trig'd it knowing it is just a 45 deg angle, thus basically immediately got r(1 + sqrt(1/2)) = 2 and thus simplified to r = 4 - 2 sqrt(2). Obviously same result as you got, just with the usual trick of expanding the fraction by the conjugate of the denominator.
Faster way: The point of the circle that touches the small square is 2 units from the right, and we have a radius going horizontally plus a radius at a 45deg diagonal to make that 2-unit horizontal, meaning 1r + r/sqrt(2) = 2 => r = 2/(1+1/sqrt(2)) = 1.17157
Radius of incircle of right triangle = (sum of sides - hypotenuse)/2
Standard result from properties of triangles
Haha, good catch.
connect diagonal from top left to bottom right, it will be tangent to the circle and form a triangle
then use formula rp=S where p is a semiperimeter, S is an area of a triangle and r is radius
you can find the area of triangle by using heron's formula and a semiperimeter by summing up three sides and dividing them by 2.
I like the dolphins behind you
How exciting
😂😂😂😂
💀
I’m with you…how exciting! When I help my teenager with math problems, I’ve started saying “how exciting” when we get to a solution. Not sure it’s working but I keep trying!
Dude he does it yet again. And yet again, it is exciting.
I also remember there was a formula to determine the radius of an inner circle of a triangle, which you could transform this problem into
Please always rationalize the denominator. You'll get rid of the fraction and you'll only have one root in your answer: 4 - 2sqrt(2) is so much nicer and simpler.
Just draw the diagonal of the big square from top left to bottom right. You get an isosceles triangle and the circle will be it's incircle. Then it's easy to calculate the radius.
This seemed like one of those questions on a math test I would skip because it took way to many steps and not enough time to deal with it when you have 3 other questions that are just as long but are easier to solve.
i took a different approach as follows:
First, i took the point of contact between the circle and the top right corner of the inner square, and make it the origin point of a X,Y coordinate system.
Then the general equation for that circle in that X,Y coordinate system would be (X-a)^2 + (Y-b)^2=r^2 where "a" and "b" are the horizontal and vertical distances from the center of the circle to the origin point. And because the circle is bisected by the diagonal of the big square, we can asume "a" and "b" are equal, and we call this value c.
Thus, the equation now comes to (X-c)^2+(Y-c)^2=r^2
Now, we know when x=0 then y=0 wich leads to r=c√2 (1)
Then, we also know that for x=c then y=2 wich leads to c=2-r (2)
Then replacing (2) in (1) we get r=(2√2)/(1+V2) ---> r=1,17
i solved it differently, i made a right angled triangle using from the center of the circle to the top right diagonal of the square. my hypotenuse was r and my side lengths were 2-r. then afterwards i quad formula the quadratic and got 4-2root2 which gives the right answer
Same
@@Eagle_SFM Not same, but similar
I love how I always end up solving it in a different way for the same answer
I did it a bit different, we can ignore the big square and just consider the right top quarter, and since the circle is limited on one side by the edges of that square and on the other side by the corner of that square, then we can draw 2 more circles:
1. draw a circle that goes thru all corners of the square, find the diameter, here ofc the diameter of the circle is the diagonal of the square, so same as in the video, using pytagora -> 2 root(2)
2. draw another circle, that fits entirely inside the square, a.k.a. inscribed in the square
-> here we can see the diameter is the length of the circle = 2
-> and we can see the difference between the diameter of these 2 new circles is 'x' on both sides, so big circle diameter = small circle diameter + 2x => 2 root(2) = 2 + 2x => x = root(2) - 1
-> we can also see that the difference between the diameter of the new smaller circle and the original circle is just 'x', so original circle diameter = small circle diameter + x => d = 2 + x => d = 2 + root(2) - 1 => d = 1 + root(2)
-> now we can find the radius by dividing the diameter with 2 => r = (1 + root(2)) / 2 = 1.2071
I went about this differently and got the same result. Draw a diagonal line from the upper left corner(call it A) to the lower right corner of the large square. Find the point of the circle that touches this diagonal (call it B). The length from the upper corner to point A is the diagonal of the square, which is 2 x sqrt(2). Find the point where the circle meets the upper side of the square (call it C). AB = AC. r = 4 - AC = 4 - 2 x sqrt(2) = 1.17.157
Why are you using Pythagoras for all those diagonals ?
It would be faster to use the formula d = s * √(2) where d is the diagonal of the square et s its side.
And where do you take this formula from?
@@user-gd9vc3wq2h It's a basic formula that you learn in high school...
Or more precisely, 4 - sqrt(8).
Co-ordinate type geometry is an easy way to see this. Centre of the circle is (x,y). We have two radii from (x,y) to (4,y) and (x,4). The radius is therefore 4-x (or 4-y), so we see x = y, as we'd expect from the symmetry of the construction.
Now we have another radius from (2,2) to (x,y). So by Pythagoras we have 2(x-2)^2 = radius^2 = (4-x)^2. Solve this to get x = sqrt(8) (clearly x>0 by construction) and therefore the radius is 4-sqrt(8).
More cleanly, 2*sqrt(2)/(1+sqrt(2)) = 4-2*sqrt(2). So... that means the diagonal between that bottom right corner and the center point of the circle is exactly 4. In case anyone was interested in a more intuitive way of looking at the problem.
That was a real pleasure to watch. Thank you!
"Punch it into a calculator." You monster!
r=4-2√2.
I guess I did this one the hard way. First calculate the diagonal of the small square, as shown. Then extend that line to the center of the circle to create a right triangle with hypotenuse 2√2 + r. Each leg is 4 - r. The Pythagorean Theorem means that (4 - r)^2 + (4 - r)^2 = (2√2 + r)^2. Multiply it out and you get a knotty quadratic: r^2 - (16 + 4√2)r + 24 = 0. This actually is amenable to solving with the quadratic formula. You get a root within a root, but that can be handled. Finally, r = [(16 + 4√2) +/- (8 + 8√2)]/2. The larger root is way too big for the given parameters. But the smaller root, 4 - 2√2, does equal 1.17157.
Now start solving jee advanced math problems 💀
Back when I was in school, my teachers would have wanted me to get the radical out of the denominator.
The radius of a circle inscribed in a triangle is two times the area of the triangle divided by the perimeter of the triangle. The area of the triangle is 8. The perimeter of the triangle is 4 + 4 + 4xsqrt(2). The radius of the circle is 16/(8+4xsqrt(2))=1.17157.
I loved multi step problems like these in school. It felt like going on a bunch of side quests in a game.
And it actually took some time to solve, which meant less time bored in class.
lol I too a completely different approach (didn’t have pen/paper so I needed an easier way):
basically, assume the center of the circle is (x,y). The center of the circle must lie on the main ascending diagonal of the square for symmetry reasons, so x=y. The lower square touches the circle at (2,2). Using this, the distance from (2,2) to (x,x) is sqrt(2)*(x-2)…
But going from the center of the circle to the top of the large square is also a radius (4-x)… you can do the same horizontally. Thus, we get sqrt(2)*(x-2)=4-x.
Notice that this is a simple linear equation in one variable… solve for x
we immediately get:
[sqrt(2)+1]*x = 4+sqrt(2)*2
x = 2[sqrt(2)+2]/[sqrt(2)+1] = 2*sqrt(2)
… therefore, radius = 4 - 2*sqrt(2) 🥳
✨as a bonus this mean’s the CZcamsrs expression 2*sqrt(2)/[1+sqrt(2)] simplifies beautifully to 4 - 2*sqrt(2)
I’m going to start doing these problems with my kids because I think if they saw how you legitimately enjoy solving these problems they could learn to do the same, and I actually enjoy math when it is approached like this. I honestly wish you all the happiness in the world brother. You’re making a difference.
I use BPT theorem to find radius of the circle by making two equal triangel with common hypotenious which is equal to the diameter of the square , and let the base of these two triangle by making one horizontal and vertical line from the side which is touching with the circle which makes the two triangle who follow the BPT rule because the length "r" both horizontal and vertical and "2units" both horzontal and vertical are prallei t each other . Now i put the values directly and indirectly and apply the BPT to get radius
thanks for giving such cute question
❤🔥❤🔥
edit= my approx anser is |1.18| because i use"root2" as 1.14
Very clever solution!
Next time: for a square with a side length of "a", the diagonal is ALWAYS "a\sqrt 2".
That is one long way to do it. Or draw a line from one corner to the other where that forms a triangle with the circle inside it, and the circle, that line (hypotenuse H), and the top right corner of the little square all touch. Then the formula for the max radius of a circle that can fit in a triangle is r=(P + B - H)/2 P and B (perpendicular and base)
r= ( 4 + 4 - 4sqrt(2) ) / 2
r = ( 8 - 4sqrt(2) ) / 2
r = 4 - 2sqrt(2)
Solved it a different way:
1. Rotated the whole thing 180 degrees
2. Now it looks like the co-ordinate system
3. One side of square as X-axis and the other as Y-axis
4. We can clearly see that the circle is passing through the point (2,2)
5. Now it's just a simple question that is taught to us "A circle is passing through a point (2,2) and touches both the axes, find the radius"
6. If we take x^2 + y^2 + 2gx + 2fy + c as the general equation of the circle, the radius become sqrt(g^2+f^2-c)
7. The circle touches both the axes, so g^2 = f^2 = c, Hence the radius = √(g^2+g^2-g^2) = g = f = √(c)
8. Put that in the equation, x^2 + y^2 + 2√cx + 2√cy + c = 0
9. Substitute the point (2,2) in the equation and take squaring on both sides. You get a quadratic and put that in calculator and you get c. Now we need √c so just sqrt the number you get and that's the answer.
Its not the fact he solves these things - I dont think that I’ve ever seen him do maths that I didnt know - but its the ingenuity with which he solves it.
Its one thing knowing how to get through the math, and another thing entirely on seeing a bunch of shapes and knowing what math to use to get you there.
I just put a diagonal in the square from the top left to the bottom right made the sides equal to n+r and the diagonal to 2n since we have 90° we know that the diagonal is equal to √(2×4²) meaning n is half of that than just 4-n and you have r. If you continue with this you can get that r is about equal to 0,29289[((2-√(2))÷2), 1-√0.5]×a side of the big square for all values if you have different side lengths it would obviously change, even tho you would still be able to generalize that.
My answer: 4-2root2
My working:
The vertical distance from the corner of the little square to the roof of the big square is 2
The vertical distance from where the circle touches the corner of the little square to the roof of the big square is radius + (radius/root(2)) since the angle is 45 degrees
Then just do the algebra for 2 = r + r/root2
I loved geometry at school for such tasks
An alternative form of the solution is 4-√8, now solve the 3D equivalent of the excercise (A half size cube inside a cube and a spehere touching its corner and the faces of the bigger cube, what is its radius?)
For any triangle S=p*r, where p - half perimeter, r - radius of inscribed circle. In our case, S = 0.5*4*4 = 8 (half the square); p = 0.5*(4+4+4*sqrt2). Thus, r = 8/(4+2*sqrt2) = 4/(2+sqrt2) = 2*(2-sqrt2) =~ 1.1715
Define your right angles when stop the problem.
My approach to this problem is different we don't need the square, divide the squared into 2 parts diagonally and this should give you a circle inscribed in a triangle then therefore we can use the formula r=(a+b-c)/2 a=b and c=4root2 r=(4+4-4root2)/2 or 1.1715...
The diameter of the circle is larger than 2. Assuming the circle is tangent to the outer box in both parts of the corner that diagonal line would never intersect the center point of the circle. so your answer is close but not correct
Big diagonal equals big diagonal. Mmm, yes. Quite.
I solved it using trigonometry. It's essentially a circle inside a right isosceles triangle.
So the radius comes to 2√2 x tan(π/8)
Can you explain how your circle comes to the centre point of the square, and has an arc that extends past the centre on two sides only?
I have a faster way, but it requires knowledge of the quadratic formula or completing the square.
Andy’s solution is more convoluted, but utilizes simpler techniques that more people will understand 👍🏼
You're making me interested in math again.
Exellent solution! I am trying to make understanding of this, and 1:30 you said you can sqaure root both sides of the equation. Unfortunately I don't think this works as 3^2 * 4^2 = 5^2, however 3 * 4 does not equal five. I would be gratefull if you could shine some insight on this.
i feel so smart watching you solve these 😎
sqrt(2)r + 2x = 2r, while 2r-x=2. x is the side length of the small rectangle between the circle and the large rectangle.
Draw a Tangent to the circle from bottom right corner of Larger Square(where tangent touches the circle at centre of the larger square) and this length is half the diagonal of larger square = 2√2. Now tangents drawn from a point outside to the circle are equal in length(one along the diagonal and one along side from the bottom corner of larger square) hence 2√2 + r = 4
r = 4 - 2√2
OR
Draw the diagonal of the larger square (connecting bottom right corner and top left corner vertices). Now we have a Right Triangle where the sides are 4, 4, 4√2 and circle is the incircle of this triangle whose radius is (4+4-4√2)/2 = 4-2√2
PS :: Inradius of a Right Triangle, whose sides are a,b,c with c being hypotenuse is (a+b-c)/2
Last advice to use calculator would lower the score when I was in school. My teachers always emphasized to use exact computation, in this case it means that fraction should be multiplied (upper and lower part) by (sqrt(2)-1). Then upper part becomes 4-2*sqrt(2) and lower part becomes 1 (because (a+b)*(a-b) = a^2-b^2, it his case a=sqrt(2), b=1 so a^2-b^2=2-1=1). Final value should be 4-2*sqrt(2).
If the goal was to solve this problem using the most complicated method possible it was a success.
It's a lot less calculation intensive to solve this problem (starting from the drawing with the 3 radii) by simply adding a diagonal line from the top left corner of the square to the center of the square which is a point of tangency with the circle.
Because we can see that this diagonal line is the hypotenuse of an isosceles right triangle with side lengths of 2, we know that this hypotenuse length will equal 2√2 and therefore the length of the square's top side left section (from the left corner to the point of tangency at the top of the circle) is also equal to 2√2.
The drawing has already established that the square's top side right section (from the point of tangency at the top of the circle to the right corner) is equal to the length of the radius.
The length of the radius equivalent section of the top side of the square (r) can be calculated from the difference between the total length of the top side of the square (4) and the length of the left section of the top side of the square (2√2).
r = 4 - 2√2
r ≈ 1.17
Diagonal of square will always be a(underroot2)
a=side of square
Mathematics is beautiful
Man, I always keep missing the little details. Thank you for the video!
My solution:
- The large square has a side of 4. Take 2 from the small square. We have now 2.
- From the circle we have r+r*cos(45°) which is equal to 2
Then r+r*cos(45°)=2
r(1+sqr(2)/2)=2 -> r≈1.1716
The top right point in the square is the director circle to the given circle now using coordinate geometry results we know distance of such point from center of circle is always r√2
Director circle : locus of points from which tangents drawn to a curve are mutually perpendicular
Where r is the radius of given
Hi sir, I highly admire the way you solve math problems it emphasizes my love toward solving complex problems. Thereby can you make a pdf (free for veiwers) based on some questions like these amd other of your math problems for guys like me to solve.
Ah I knew it would have to have something to do with triangles, but couldn't figure out exactly how!
Logic flaw here, if you rotate side 'c' down onto side side it 'must be longer' than side 'a', if it were actually 2 units the angle of 'ca' would be zero or side 'a'. Simple proof on paper, create the box, draw the diagonal 'c', get a protactor (Lead pencil) spead the legs the distance of 'a' and rotate the protractor to the verticle, 'c' should be bisected.
Caou say hat in German?
You should have rationalized the denominator by multiplying top and bottom by (1 - sqrt(2)). The denominator simplifies to 1, which eliminaties the fraction completely, and the numerator becomes 4 - 2(sqrt(2)). Evaluating that gives the same answer: 1.17157...etc.
Also, much simpler way to do this: Extend the horizontal top of the lower left square to the right-hand vertical side, dividing it into 2 segments, each length 2 (just like the left side). The top part of the top segment is r, as you've shown. The bottom part of that top segment is the side of a square whose diagonal is r. That's r/sqrt(2). So now we have r + r/sqrt(2) = 2. Solving this equation gives the same result.
Nice solution! If you want to find the exact answer, then multiply the denominator by the conjugate. (2√2 - 4)/3.
Double-check your denominator there, I think you added the squares instead of subtracted.
I think he is making this things up. This is crazy!
I think he should have taken a different variable name apart form c for finding the hypotenuse using the circle with radius r as he had already used variable c for the square of sides 2
Well.. to find the diagonal. we can directly use the formula. Side×√2
Here side = 2, so diagonal is 2√2.
I suck at math but unless I am mistaken the radius of that circle has to be half the distance from the connecting corner of the lower right square to the square to the opposite corner of the upper right square and that is simply the square root of 2 squared + 2 squared = 2.828
no?
As a draftsman and tinkerer, it was pretty immediately obvious that r+r/sqrt2=2.
I just want to know how or where he gets these problems. They really are exciting.
I find the abswer.
first make a triangle by connecting diagonal
area of triangle is 4 x 2 = 8
the perimeter of triangle is 4√2 + 8, so S = 2√2 + 4
R = 8 : (2√2 + 4) = 4/ ( √2+2)
Along the diagonal we have
r . (1+sqrt(2)) = 2.sqrt(2)
Or
r . (1+1/sqrt(2)) = 2
Or
r = 2/(1+1/sqrt(2))
At the end multiply the top and bottom of that fraction by (√2 - 1) to clean up the answer