Circle around squares solved in 3 ways
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- čas přidán 2. 08. 2024
- Thanks to Charlie from Australia for this suggestion! The problem comes from the Australian Maths Competition (AMC). A circle circumscribes 5 squares as shown. What is the radius of the circle?
0:00 Problem
0:43 Method 1
4:10 Method 2
6:24 Method 3
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I really liked the multiple methods format, I often solve your problems but rarely in the same way you do, this time my method was included. Plus it's really cool that in maths all the answers have to agree no matter how you find them and this format highlights that.
Here is another method: Fitting the circle formula.
Apply coordinate axes as in method 3 in the video (at 6:30), with horizontal x-axis, vertical y-axis, and O = (0,0) at the bottom-left vertex of the top-left square. We know coordinates of the circle at (0,1), (0,-1), (3, 0.5) and (3, -0.5). Now fit the general formula of a circle through those points:
(x - a)² + (y - b)² = r²
where (a,b) are the coordinates of the circle's center, and r is the radius of the circle.
Since we know the center of the circle is on the x-axis, we already know that b=0 . So now we have to fit only (x,y) = (0,1) and (x,y) = (3, 0.5) into this formula; plugging in these coordinates, we get:
(0 - a)² + (1 - 0)² = r²
(3 - a)² + (0.5 - 0)² = r²
Expand parentheses:
a² + 1 = r²
9 - 6a + a² + 0.25 = r²
Subtract first equation from second equation:
8 - 6a + 0.25 = 0
6a = 8.25 = 33/4
a = 11/8
Entering this result into a² + 1 = r² :
(11/8)² + 1 = r²
r² = 121/64 + 1 = 185/64
r = √(185/64) = (√185)/8
_E presto!_
I did it the same way
@@prabhatsitaula2873 So did I!
I did something similar. I put the origin at the centre of the circle, which I made an unknown distance b to the right of the centre of the 4 squares. Solving for b and r was easy, using the coords of two points on the circle. It was much quicker than Presh's method.
@@garethb1961 Actually, that calculation is the same to how I first solved it (see my other comment) and corresponds to Presh's method 1 (the "right triangles" approach), except he defined triangle's horizontal side length = x , while you and I defined triangle's horizontal side length = 1+b (or 1+x) .
In other words, we'd have two right triangles with Pythagorean Theorem applied to them:
(1+b)² + (1)² = r²
(2-b)² + (½)² = r²
Solve for b and r .
My favorite method.
Nice video Method 3 was the best. I would be very happy if this was taught in schools
It is taught in schools using that method only .... !
Why would you say that method is "best"?
By what criteria did you decide that?
(I ask in part because, to be honest, I disagree. I find that method the least imaginative and thus discount it. I discount method 2 because it requires the most "jargon" or knowledge. Method 1 seems simple and logical, and - by THOSE criteria - I would deem it the best.
However, those criteria are simply my own.
I would NOT claim that it "IS" the best, but merely "my" best.)
So: what defines "best" for YOU?
@@araptuga The coordinate method is the most general, can be applied to the most problems, and can come to your rescue when other methods fail. Also, the fact that it requires no imagination can be considered a strength of the method. It can pretty much be automated. That, to me, has a beauty all its own.
Thanks for that explanation. Obviously "best" is a subjective choice; nobody's right or wrong. But it can still be a useful description, as long as it's understood under what "rules" it's being judged.
I have solved this by,
First calculating any of the 3 vertex of those squares which passes through the circle ( we have 4 so we can choose any of the 3)
Then just use the equation of the circle
Put those points in the equation and we will get the r
Love the multiple solutions. I used the first one but loved the coordinate solution. The cartesian plane is such a powerful tool, no wonder it created a revolution in mathematics and physics.
I agree Descartes for me !
The 2nd solution is easiest if you remember the formula.
I solved it using the method 1 and it was my 1st question of your video I solved without a minor hint from video😊
Another method:
Consider the triangle ABC shown at 4:38. Let the line DC splits the angle C into C1(upper) and C2(lower). C = C1 + C2. Apply tan formula for sum of angles.
tan C = (tanC1 + tan C2) / (1 - tanC1 tanC2)
= (05/3) + (1.5/3) / (1 - (05/3) . (1.5/3))
= 8/11
Let the center of the circle be O. Then angle ACB will be half of angle AOB. Also, since the triangle AOB is isosceles, the angle AOE (let E be the mid point of AB) will be same as ACB.
tan of angle AOE = 8/11 = AE/EO => EO = 11/8
Now, radius = √ (1² + (11/8)²
This extended another approach where you can extend the horizontal line from point A and let it cut the circle at C' then angle C' = angle C. And as we know upper left most is a square so angle BAC' is 90 degree which would make like BC' the hypotenuse and also the diameter of the circle. AB = 2, tan C' = 11/8 so AC' = 11/4 => BC' = sqrt(4 + 121/16) = sqrt(185)/4
and as BC' is diameter so radius is sqrt(185)/8
I thought the last solution using coordinate geometry was the most intuitive and the simplest, but I love that you included multiple methods here! 👍
By the intersecting chords theorem: ab / c = d so (3/2) (1/2) / 3 = 1/4.
c + d = 13/4 and side of a square = 1 ⇒ r = √((13/4)² + (4/4)²) / 2 = √185 / 8.
Alternatively, the formula r² = ¼ (a² + b² + c² + d²) could also be used.
I used coordinate geometry and subbed three known points into the formula for a circle, (x-h)^2 + (y-k)^2 = r^2. Then I used simultaneous equations to calculate h and k, followed by subbing in another point to get r.
Nice one. I did it using method 1, then a hybrid of 2 and 3...... I like these problems, nice brain exercise that I can actually do!
Excellent.
These were your most difficult.
I loved the 3rd solution, many people don't like to go into analytic geometry, but this one was really well presented! ♥
Mine was similar to the second method, except after finding the three side lengths, I used the law of cosines to get the angle at point C to be arccos(11/sqrt(185)). Then, using the inscribed angle theorem, the central angle AOB is twice this, but is bisected by the horizontal line. Thus the angle AOM, where M is the midpoint of AB, is exactly arccos(11/sqrt(185)), and using basic trig on the right triangle AOM, with AM = 1, gives the radius OA = sqrt(185)/8.
I did it very similar to the 3rd way. Didn't think that would be included as it is a strange way to solve it but I was happy when I saw a similar method to mine. Love the vids, keep it up! 😊
Good problem and nice solutions! I solved this one in a different way than all 3 given solutions. Since I'm pretty weak in geometry, my solution mainly involves algebra. I first set up a coordinate system centered at the center of the 4 squares on the left and used the x-coordinate of the center of the circle as an unknown, x0, yielding the equation (x - x0)^2 + y^2 = r^2. I then plugged in the known values of x and y for the two points (-1, 1) and (2, 1/2) lying on the circle and solved the system for x0 and r, yielding x0 = 3/8 and r = sqrt(185)/8.
I also used coordinate geometry with the same coordinate system that Presh chose. Except that I used the formula for a circle, namely (x-a)^2 + (y-b)^2 = r^2, where (a, b) is the center of the circle. I used that to get 3 equations by substituting 3 x,y values [namely, (0,1), (0,-1) and (3,1/2)] to get 3 equations, which are then solved to get the values of a, b, and r^2: a = 11/8, b = 0, r^2 = 185/64.
I have the easiest method to solve the problem by using the chord intersection formula and using semicircle . First see the right side square and top corner with up side and extend the side upto left side circle boundary and we find a semicircle with right and triangle which side is 1 and (3+x) and diameter .and for value of x we using chord intersection by (1.5×0.5=3×x) then find value of x =0.25 and now semicircle base 1 and perpendicular 3.25 then diameter found is 3.4 so radius is 1.7
I solve this problem without using pen copy only 30 second see the figures and calculate the easy way and also radius .
Please read my text which I have writen.and if I'm right then I want like by Presh Talwalker I'm best in math
I also qualified iit advance I want some tips for become a mathematician and also ask more thing
that is a good exercise ! all 3 methods were educational and useful for implementation , you can also solve this using only thales and pythagore , as always thank you for this juicy geometry exercise !
There is also the 4th option by using the sine law a/sin(alpha)=2R
R=1/(2sin(atan(0.5)-atan(1/6))) = 1.700
I initially did it with not the most elegant method:
Solved for the hypotenuse of a right triangle with base and height 2 (within the 2x2 grid of squares), i.e. 2^2+2^2=c^2, with c being 2sqrt(2) in this case (positive for the sake of example).
The solved for half of hypotenuse of right triangle within one square, i.e. the hypotenuse of a square with base and height 0.5 since I was focusing on the far right square with the remaining gap. 0.5^2+0.5^2=c^2, c = 1/sqrt(2), (again positive for the sake of the example).
The combining these terms, 2sqrt(2) and 1/sqrt(2), we get 5/sqrt(2). This of course would be a chord (which is very close to the diameter in approximation) so our final answer would be approx. 5/2sqrt(2), which is approx. 1.76...
So I was pretty close for using a bunch of educated guessing.. (being ~0.06 off)
Edit: of course considering the number I found would be half of a chord very close to the diameter, it could be slightly closer or further from the correct answer, but close nonetheless for an elementary solution lol
Yet another analytic-geometric method is to compute a 4x4 determinant. The first row of the matrix is (x^2+y^2,x,y,1), while the remaining 3 rows are the results of substituting the coordinates of 3 points on the circle into that first row. Setting the determinant of that matrix equal to 0 gives you an equation of the circle through those three points, which is then easily manipulated to extract the radius. Clever choice of the coordinate origin can make that determinant very easy to compute.
Sorry, but I'm not familiar with such a method so I don't know how that works. Can you show it more concretely?
What does (x^2 + y^2) represent? How do we substitute the coordinates of three points (for example, (0,1), (0, -1), (3, 0.5) , using the coordinate axes as in method 3 in the video)?
EDIT: By the way, I do know how to evaluate the determinant of a matrix , but it's how to fill the matrix that's puzzling me.
Yay! Solved it your first method and nailed it :) Also considered method 2. The coord method was fun...
As I watch this channel I'm getting a better appreciation on how hard problems are to solve. The hardest to solve tend to be those that have only one way of doing it, and the method involves some non obvious (to most problem solvers) trick. The easiest problems tend to have more than one way of solving. When Presh starts out by "there are many ways to solve this problem" you know it's not very challenging to most subscribers to this channel.
Probably the hardest problem I've seen on this channel that presents itself very simply is the ladder leaning against a cube box and touching the wall. The set up was simple but Presh presented a very complicated solution and there didn't seem to be an easier way to do it.
Method 3, I must add this to my 'tool box'; very simple yet powerful.
Loved coordinate geometry in school. 3rd method is so much fun.....
Thanks 🙏
This is why I am so hooked on maths.
I used the ratio of 2 intersecting cords twice to get the diameter and then just divide by 2
You can also find the equation of the perpendicular bisector of AC and another equation (equation of x axis i.e. y=0). you can then solve these equations, which will give u the coordinates of the centre.Then, using distance formulae the radius can be found
I really like you videos
I really like videos too, wouldn't personify them tho
I like the video. However, there is one thing missing. I say this because in college we discussed mathematical rigor which in part means you have to be aware of assumptions you make.
Since three points determine a circle, you aren’t guaranteed that you can circumscribe a circle through the four given points. In this case you can. Just be aware that because you drew it, does not always mean it can actually exist. There are good examples where a drawing looks correct, but leads to an incorrect answer because of the assumptions made from the drawing.
*My (4th) Method*
Use extended sine rule - a/sinA = b/sinB = c/sinC = 2R (diameter of circum-circle), where origin = center of big square and A = (-1,1), B = (-1,-1) and C = (2,1/2)
c = 1 - (-1) = 1+1 = 2, so 2R = c / sinC = 2/sinC, therefore we have our answer R = cosec C. *Easy so far, Right* ? Use vector dot product to calculate cosC since both CA and CB vectors are computable from the co-ordinates of A,B,C.
CA = A - C = (-1,1) - (2,1/2) = (-3, 1/2), so length = sqrt(9+1/4) = sqrt(37)/2
CB = B - C = (-1,-1) - (2,1/2) = (-3, -3/2), so length = sqrt(9 + 9/4) = sqrt(45)/2
CA.CB = dot product = x1 x2 + y1 y2 = -3 x -3 + 1/2 x -3/2 = 9 - 3/4 = 33/4
So, cosC = dot product of unit vectors = (33/4) / ((sqrt(37x45)/4) = (3 x 11) / (3 x sqrt(37x5) ) = 11/sqrt(185)
So, sin^2 C = 1 - cos^2 C = 1 - 121/185 = 64/185 = 8^2 / 185
Hence, circum-radius = R = cosec C = sqrt(185)/8
Hey Presh, I read on your about me page that you studied mathematics and economics at Stanford, and I was thinking about doing the same. Do you have any advice for me? As always, thanks for the amazing content you produce. Peace and love.
1:54
Yeah I also want this same advice...
Get your education and create a youtube channel doing the same thing. Except, tailor it to real world solutions, like architecture, not just math problems. Also, make them interesting. A sniper sits on a building 500 ft high. His victim, who is approximately 5'6"tall is jogging at 10 mph, due east. A constant breeze is blowing from the west at 15 mph. The rifle fires a bullet at 1500 meters/second. Where would the sniper have to aim to intercept the jogger in the head? (Doesn't have to be gruesome like this, but creative to keep people watching videos)
Pythagorean theorem: D= sqrt(1.5^2+3^2)= sqrt(11.25) then divide by 2 for radius. R= (1/2)*sqrt(11.25)
It’s embarrassing that I missed this! I’m in calculus 3 and differential equations!
You calculated the length of a line segment that is _not_ the diameter of the circle (nor is it the radius; note that sqrt(11.25) > 3 ).
Actually this is how one finds 2r 😉
@@dprxsv If you multiply this result by √(37/36) , then yeah...
@@dprxsv ah crap you’re right
Where do they teach the circumradius formula again? It is new to me. I am glad I had method 1 and can understand method 3, at least.
Never heard it before either. It seems like such a completely random formula
One of the first few
I ended up put the whole thing into X-Y chart but I make (0,0) the middle of the second square column so, the left points become (-1.5,1) and (-1.5,-1) and the right points become (1.5,0.5) and (1.5,-0.5)
then I found the center of the circle to be (-0.125,0), after which I put all value to gather and get that same answer
Would it be possible to solve it using trigonometry?
I can never remember formulas, so I almost always solve these types of problems by finding the coordinates of the circumcenter of the triangle ABC. However, I would usually find the equation of the perpendicular bisector of AC by converting from point-slope, y - 3/4 = 6(x - 3/2), to slope-intercept, y = 6x - 33/4, to find the intersection with y = 0, which is the perpendicular bisector of AB by construction:
y = 0
y = 6x - 33/4
6x - 33/4 = 0; x = 33/24; p = (33/24, 0)
Then apply the distance formula to find r. The right triangle method is clever.
pretty easy
Here's a similar problem. Consider the 4 squares on the left, and an equilateral triangle with edge length on the right. Find the radius of the circle that circumscribes the two square corners on the left and the rightmost vertex of the equilateral triangle.
What's the length of the side of the equilateral triangle? Is it 1 , or is it 2 ?
If it's 1 : (using a "right triangles" method like method 1 in the video)
1² + (2-x)² = r²
x + (√3)/2 = r
1 + 4 - 4x + x² = r²
x² + 2x*(√3)/2 + 3/4 = r²
5 - 4x + x² = r²
x² + x*√3 + 3/4 = r²
Subtract first equation from second equation:
4x + x*√3 + 3/4 - 5 = 0
x(4 + √3) = 4.25 = 17/4
x(4 + √3)(4 - √3) = (17/4)(4 - √3)
x(16 - 3) = (17/4)(4 - √3)
x = (68 - 17√3)/52
Insert this result into x + (√3)/2 = r :
r = (68 - 17√3)/52 + (√3)/2 = (68 + 9√3)/52
If the triangle's side has length 2 , then the equations become:
1² + (2-x)² = r²
x + (√3) = r
and via a similar calculation we eventually arrive at r = 2 .
Same as mine.
Good evening
Please please Mr Presh Talwalkar, take a look on the net about a false demonstration of the equation ''1^x=2'' which have no solution even in C, but the demonstration is well done that many people can believe at.
Thank you very much.
Ali, from Tunisia
I using method no 2
2 method
the second method is the easiest, in the third it is easier to write three equations of a circle
My favorite method to do this is measuring tape
Solution:
M = middle point of the circle,
x = distance from M to the right side of the right square.
Pythagoras on the right side:
(1) x²+0,5² = r²
Pythagoras on the left side:
(2) (3-x)²+1² = r² | (1) = (2) ⟹
(3) x²+0,5² = (3-x)²+1² ⟹
(3a) x²+0,25 = 9-6x+x²+1 |-x²-0,25+6x ⟹
(3b) 6x = 9,75 |/6 ⟹
(3c) x = 1,625 |in (1) ⟹
(1a) r² = 1,625²+0,5² = 2,890625 |√()
(1b) r = √(2,890625) ≈ 1,7002
Sir I just round off my answer from 1.7 to 2 so will it work ? meaning is 2 as an answer acceptable?
Also, in method C, you say that the perpendicular bisector of AC passes through the center of the circle. That's not an assumption, right? On what basis (a theorem, rule) do you make that conclusion ? Thanks in advance. I find these math questions interesting but are really beyond what little is left in my brain from my high school math :)
Perpendicular bisector of AC is set of all points equally distanced from A and C. On the other hand A and C as points on a circle are equally distanced from the center of the circle. It means that the center of the circle is one of the points equally distanced from A and C, or in other words, element of perpendicular bisector of chord AC defined by them. :o)
For method 2, why didn't you make the triangle an isosceles with point C being in the middle of the rightmost vertical edge of the rightmost square?
Point C then isn't on the circle, so how would you proceed from there?
How did he got 3 in the 1st method ?
the width of 3 squares is 3 by definition. With that in mind, if a part of this total distance of 3 is x, the rest has to be 3-x.
In method 2, on what basis do you conclude that AD is exactly 0.5 ? I know it looks like it is bisecting the side of the upper square, but how do you theoretically prove that without a drawing done to scale?? Pls explain, thanks in advance.
It is given that the fifth square is at the middle of the 2x2 grid. Therefore C is half unit up from the horizontal centerline, while A is full unit up.
we have the trangle we have everything obviously
The statement at 1:32 needs demonstration.
I also have a good question suggestion for you...
Where to send..?
To your given Mail..??
i did it by constructing a triangle of which the hypoteneuse passes through the upper 2 boxes and the side one just like method 2 and then assumed that to be the diameter of the circle. i still got 1.7 as the radius. is my thinking right?
No. The length of that line segment is √( 3² + 1.5² ) = √( 9 + 2.25 ) = √(45/4) = (3√5)/2 = (6 * √5)/4 , which is not the diameter of the circle because the actual diameter is (√185)/4 = (√37 * √5)/4 . So it's close (in this case; only about 1.4% off), but not correct.
@@yurenchu thnks for your reply.
Your videos are so inspiring that we can solve them using the simple and clever method within a few seconds, tbh!❤👀 Keep posting challenging and intriguing problems for our benefits 😊😊
2:36 how it is 3-x??
The whole segment is 3 unit squares long. If x is distance of P from the left end then distance to the right end is 3-x.
For method 1, would have been simpler if you defined x and 3-x as the opposite distances.
It doesn't make a difference; you'd expand parentheses of the same terms, and you'd end up with a similar linear equation in x after subtracting the two equations from eachother.
I calculated 1.8 in my head just by visualisation
I used law of cosines.
oh hey... I guess 2 out of 3, not bad
Isn't it sad that we waste away for 10-13+ years of school not knowing why we do maths and what to do with maths? And then some of us become artists of any kind and are also perfectionists and suddenly we need to do so much maths! And we could have paid attention is school, but it's impossible to retain everything, so now we need to figure it out by ourselves and then you realise: If I'm gonna figure it out by myself as an adult anyway, why did I even need to learn this in school? Or: Why didn't they teach me what this can be useful for and support my hobbies and interests sooner?
So this problem actually has real world application but I'd just pop it into a CAD program and tell it to tell me what diameter circle I need. Much faster than doing this by hand
r is at least 1.5. I'll guess 1.7
haven't seen the video at all, the answer is Pi. I win.
fu-
Height of M in last solution should be proven.
Like if you solved it in your head from the thumbnail 🤣🤣🤣
used the 2nd method. Have to use pencil and paper though:)
I got the answer by a 4th method....
To me it wasn't clear that half of the cube was above the middle line and half of it below. Could have been 0.6 and 0.4 too.
also at 5:00 how do you know it is 0.5 and 0.5 (exactly half)?
He said that the fifth square is at the middle of the 2x2 grid when he defined the problem.
@@curiousobserver5381 thank you
Too easy with intersecting chords theorem, why do it the hard way ?
I dont see why anyone would.ever think of method 2..surely it'd be because you saw something similar. Ive never heard of the circumradius formula..how and why would anyone come to discover or deduce this. And Presh why don'tyou show hiw it was deduced in this video?.I see no reason anyone would think k of drawing a trial gle like that..can anyone else..drawing smaller right triangles or other tria goes within the squares yes,but doesn't everyone else think this method is out of nowhere?
Hello this is Sayed Yousaf from Afghanistan.
I found a difficult math question.
If you help and solve me it would be your pleasure.
The question is:
The limit x approaches 0 (x^x...^x-x!)/(x!^x!-1)
Congrats to all who is early and found this comment 🥳