From my experience a lot of techniques that Andy applies, and the way he applies them in the problems he solves, have genuine practical research and engineering value.
Best part is he shows all his work, even if it’s kinda unnecessary he still shows us the working even down to the most basic multiplication. So anybody can follow along
@@bobtheblob2770 For those who have a CAD package and the ability to use it. What Andy is doing is educating people on the applications of the fundamentals. I hope your computer never fails.
Math is a subset of formal logic, and thus philosophy, using very precise symbols, definitions and minimal axioms. I'm an industrial electrician (sparkmonkey) that translates philosophy on the side. I understand how you feel, and I can offer you this tentative answer: you want to learn something new and true about reality
I watch math channels on CZcams because I love math and I like to keep my memory fresh and ready to solve problems. But Andy's videos are really exciting. It makes you fall in love with solving math problems even if you don't like math. How Exciting!!
that could have worked as squares of the side lengths give you the area of the square and would have got you straight to 36 without first calculating 6.
You can also solve one of the angles of the blue triangle and then use trig functions to find the side length of the pink triangle, but your method of using pythagoras theorem was more elegant! 👍
My memory is spotted so I missed some of the shortcuts related to alternate interior angles and sorta had to do everything from scratch with basic trigonometry and actually work out all the angles. I calculated the hypotenuse of the blue triangle using plain old Pythagorean Theorem. With all the sides and one angle known, I could use sine/cosine for the remaining angles. For the top right angle of the pink triangle, I knew it had to equal 90 minus the top right angle of the blue triangle. With two out of three angles of the pink known (it's a right triangle), we just subtract the sum of the two knowns from 180 to find the last angle. With all angles and one side known of the pink triangle, I opted to use the Law of Sines to find the base of the pink triangle which would also equal the base of the pink square.
I was never much for math, but I do love a logic puzzle. Your channel reminds me that math is just a puzzle and I love that. It's been 20 years since I learned some of this stuff. Some of it I'm not sure I was ever taught. Many of your leaps go over my head, but I'm certain with extended exposure, I will begin to pick up the pattern of your logical deductions.
Andy, I'm really enjoying these puzzles. Just found them and I've gone through a few; subscribed with alerts. I'm out of practice and you're helping me limber up my math brain, thank you!
I saw your video with roots and x=69, but I was curious of your content and I saw you have a lot of videos and a website with even college complexity, you made shockingly good impression on me. Even the way you talk and move tells me you’re passionate with it. You’re a great guy. Thank you I’ll be studying differentials with you. Peace!
I’m learning so much from this. Haven’t been doing too good in geometry lately and these videos have really made me think about it more (in a positive way)
I used to love math a lot during my high school years, but due to limitation of job here in my country, I further my study in accounting instead. Watching your videos bring lots of good sweet memories. Thank you.
I am absolutely satisfied to find this channel, because I gain two benefits- learning English and learning mathematics, Just thank you very much for creating this channel. 🙏🏻
nice, I did it differently using trigonometry, yours seems neater. I realised that all the right angle triangles made by extending the lines of the squares of the pink square are the same ratio and did the maths. they square of the hypotenuse of the blue triangle was 25 (side is 5) so all that needed to be done was calculate the short side of the small triangle created by extending the pink square's right side down, which turned out to be 1. 5+1=6 is then the side of the pinks quare, so Area is 36.
For a first time in a while, I've actually decided to solve the problem myself, and at some points I thought "I might be doing something wrong" (which actually happened at proportion, oops 😬), but it was so much fun to see that you were actually on the right path the whole time
I did it a little differently: Since the 20-square is exactly 4 times the area of the 5-square, we can cut it up into four squares, each of area 5. The side length of each 5-square is clearly √5. So the side length of the big square is 3√5. The small white triangle is similar to the pink triangle, as Andy showed (right triangles with alternate interior angles). So the ratio of the long leg to the short leg in both triangles must be the same. In the white triangle, the long leg is 2√5 and the short leg is √5. So the ratio is 2:1. The long leg of the pink triangle is s (the side of the pink square), so the short leg must be s/2. The hypotenuse of the pink triangle is the side length of the big square which we already know is 3√5. So, by the pythagorean theorem, s² + (s/2)² = (3√5)² . (3√5)² = 45, so the equation simplifes to 5s²/4 = 45. Solving for s², we get (4/5) x 45 = 36. Finished; no need to find s, itself.
I got this one! Not the numbers, but the principle. I got there through the way you prove that line theorem. Top corner blue (a)+top corner pink(b) = 90, they're both right triangles, so angle sum of 180, 90 for the right angle, so the 90 left goes to a and b
Nice video. I solved the problem in a similar manner: Note that the side lengths of the three black squares are sqrt(5) times 1, 2 and 3. Therefore the short catheters of all the similar triangles are half the size of the corresponding long catheters. And the hypotenuses are sqrt(5) times the short catheters. Then the short catheter of the pink triangle must be 3, and thus the long catheter is 6, and the pink square has area 36.
The first blue triangle has ratio 1:2:sqrt(5) and is congruent with the first pink triangle, of which the hypotenusa is 3sqrt(5). Hence the pink square has side 6, and thus an area of 36.
Pretty much took those steps. One thing I noticed: In one of the last steps you cross-multiplied both sides, only to divide by one of the terms immediately after. Given that you only needed to multiply by one divisor to get the variable alone that "multiply-divide" step is unnecessary.
I solved it by using alpha = arctan(/sqrt(5) / /sqrt(20)), then s = cos(alpha) * (/sqrt(5)+/sqrt(20)) and then area = s^2 but your method was also cool
I constantly solve problems with a solution different from how the teacher solves them. This task is no exception. I found that the side of the square (which is the hypotenuse of the triangle shaded in red) and it is equal to √45, the drawing clearly shows that the smaller catheter of the triangle shaded in red is 1/2 of the second Let the small catheter be x, then the second one will be 2x. So we get a triangle with x and 2x catheter whose hypotenuse is equal to √45. According to the Pythagorean theorem, C^2=A^2+B^2, 45=x^2+4x^2 => 45=5x^2 | :5, => 9= x^2, x=3 (x is a small catheter, a larger catheter, which is also the side of the desired square 2 times larger, that is, the side is 6, the area is 36 sq units). The whole solution is based only on knowledge of the Pythagorean theorem.
I was thinking of using trigonometry to find the angles in the blue triangle and back again to get the relevant side of it's red counterpart... Much more elegant and efficient your way ^^ However for the length of the bigger square's size, I found calculating the total area and square rooting it to be faster more satisfying than the way you did it, even if it's not faster ^^ You easily find that the area is 45, meaning sides are sqrt(9*5), which is 3sqrt(5) ^^
This one feels conceptually pretty straightforward if you imagine a coordinate plane over the shapes. We know the slope of both bottom line segments on the square, and could calculate the intersection of those lines. But it does seem like some hairy algebra.
Sidelength of the given squares are √5 and √20 = 2√5. I called s = sidelength of the pink square and e = hypotenuse of right triangle on top of the 20-square. Then e = √[(2√5)² + (√5)²] = √[20 + 5] = 5. Due to similarity of triangles we have (2√5)/e = s/(3√5), hence s = (2√5)·(3√5)/5 = 6·5/5 = 6, therefore area = 6² = 36 square units.
I solved it by setting a system of equations with what he had as the pink triangle netting the first equation: a^2+(b+5)^2=3sqrt(5) and a second triangle in the top left of the big box that determined the diagonal distance must be sqrt(10), so the second equation was: a^2+b^2=10 not necessarily a better way but brings more information about the dimensions of the pink square once you solve the system of equations (either a=3 or -3 overlapping satisfies) :)
Lovely how clean the solution ends up being, no roots or fractions at all. Not having the time to do this myself right now, but I wonder if the identity works out to square 1: x square 2: x^2 - x square 3: (x + 1)^2 ? So a 6, 30, would end up being 49? Or maybe just these numbers are special…
Work was slow and I did have time actually. This does not work out :( smaller squares area 3 and 6 make a larger square with area 6+4√2 instead of my presumed area of 16 Edit: I did find the cool extension of the pretty continuation though! The sides of the large square work out to be (x-1)+√(x-1), so the “clean” solutions end up being when the smallest squares area is exactly 1 more than a perfect square, so 10 and 90 == 144 17 and 272 == 400 26 and 650 == 900 …
You could just calculate the sides of the 5 and 20 square then the side of the white square and then you reverse Pythagoras to find out the right triangle sides which have also pink square side
In India, we learn algebra like this- when any operator on one side goes to the other side of the equation, it changes to the opposite operator. Like + to -, * to / and vice versa. I think it’s a bit faster that doing it on both sides.
there is an easier way to do this because radical 20 simplifies down to 2 radical 5 so you can add the radicals to get 3 radical five and square that for the area. please tell me if im wrong.
From my experience a lot of techniques that Andy applies, and the way he applies them in the problems he solves, have genuine practical research and engineering value.
Best part is he shows all his work, even if it’s kinda unnecessary he still shows us the working even down to the most basic multiplication. So anybody can follow along
This is the kind of problem you can solve in cad in like 30 seconds without thinking
@@bobtheblob2770 For those who have a CAD package and the ability to use it. What Andy is doing is educating people on the applications of the fundamentals. I hope your computer never fails.
@@bobtheblob2770 Yeah, but thinking is good for your brain 😅
@@bobtheblob2770you will never go far with that attitude😂
"Let's put a box around it"
Always my favorite part
My favorite part is when he says "how exciting 😃"
@@spooks188 That's a close second 😁
How. Exciting.
Why do I keep watching these!? I haven't been in a math class in 12 years.. but it's all coming back. ..
Math is a subset of formal logic, and thus philosophy, using very precise symbols, definitions and minimal axioms.
I'm an industrial electrician (sparkmonkey) that translates philosophy on the side. I understand how you feel, and I can offer you this tentative answer: you want to learn something new and true about reality
Same here
These videos teach me how stupid I am.
"An intelligent man is one who knows that he knows nothing"
-Plato or some shit, idk
"All I know is that I know nothing"*
-Socrates*
I watch math channels on CZcams because I love math and I like to keep my memory fresh and ready to solve problems. But Andy's videos are really exciting. It makes you fall in love with solving math problems even if you don't like math.
How Exciting!!
Man, trying this puzzle I interpreted the 5 and 20 as the side lengths of the squares instead of the areas. Time to go to bed
same 😅, i gotta wake up tomorrow 6am for class, idk what am i doing here at night time lol
that could have worked as squares of the side lengths give you the area of the square and would have got you straight to 36 without first calculating 6.
I did the opposite
I love how he is getting happier when he is getting nearer to the result
Cool problem. How exciting
I've seen a handful of your videos now. I really appreciate how you break down all the steps and explain everything along the way. Thank you.
You can also solve one of the angles of the blue triangle and then use trig functions to find the side length of the pink triangle, but your method of using pythagoras theorem was more elegant! 👍
I did it like this too!
My deep memory of geometry kicked in when you brought up those theorems.
I thought I was doing so good until I got stuck at the alternate interior angles, which I completely forgot about.
My memory is spotted so I missed some of the shortcuts related to alternate interior angles and sorta had to do everything from scratch with basic trigonometry and actually work out all the angles. I calculated the hypotenuse of the blue triangle using plain old Pythagorean Theorem. With all the sides and one angle known, I could use sine/cosine for the remaining angles. For the top right angle of the pink triangle, I knew it had to equal 90 minus the top right angle of the blue triangle. With two out of three angles of the pink known (it's a right triangle), we just subtract the sum of the two knowns from 180 to find the last angle. With all angles and one side known of the pink triangle, I opted to use the Law of Sines to find the base of the pink triangle which would also equal the base of the pink square.
I was never much for math, but I do love a logic puzzle. Your channel reminds me that math is just a puzzle and I love that. It's been 20 years since I learned some of this stuff. Some of it I'm not sure I was ever taught. Many of your leaps go over my head, but I'm certain with extended exposure, I will begin to pick up the pattern of your logical deductions.
Andy, I'm really enjoying these puzzles. Just found them and I've gone through a few; subscribed with alerts. I'm out of practice and you're helping me limber up my math brain, thank you!
bro is making maths a game
wish i had a teacher like him
Agree👇
Andy I just discovered your channel and just wanted to say thank you for making fun and engaging math videos!
I saw your video with roots and x=69, but I was curious of your content and I saw you have a lot of videos and a website with even college complexity, you made shockingly good impression on me. Even the way you talk and move tells me you’re passionate with it. You’re a great guy. Thank you I’ll be studying differentials with you. Peace!
Finally a question that contains my knowledge about Transversals
I’m learning so much from this. Haven’t been doing too good in geometry lately and these videos have really made me think about it more (in a positive way)
Andy's video notification makes me excited
How exciting
As a 8th grade stundent from Turkey, this is lot easier than our highschool entrance exam questions.
man i remember doing problems like this in 7th grade... it was a fun time
Its amazing how your channel popped off in the last few months from 20k to 100k
Cool problem, did not expect the answer to be an integer. As always, straight to the point 👍🏻
I used to love math a lot during my high school years, but due to limitation of job here in my country, I further my study in accounting instead. Watching your videos bring lots of good sweet memories. Thank you.
انت مصري؟
I am absolutely satisfied to find this channel, because I gain two benefits- learning English and learning mathematics, Just thank you very much for creating this channel. 🙏🏻
nice, I did it differently using trigonometry, yours seems neater. I realised that all the right angle triangles made by extending the lines of the squares of the pink square are the same ratio and did the maths. they square of the hypotenuse of the blue triangle was 25 (side is 5) so all that needed to be done was calculate the short side of the small triangle created by extending the pink square's right side down, which turned out to be 1. 5+1=6 is then the side of the pinks quare, so Area is 36.
“please sir, may I have some more[mathematical problems solved in a particular way]”
Didn't even think this was possible. Incredible
How exciting to hv a teacher like u
you had me on the edge on my seat through that one 🥰🥰🥰🥰
"How exciting"
love this dude 😅😂
These are exciting and fun for me, too. Thanks.
this videos are so fun. Thanks so much.
I LOVE YOUR CONTENT MAN
How exciting!
For a first time in a while, I've actually decided to solve the problem myself, and at some points I thought "I might be doing something wrong" (which actually happened at proportion, oops 😬), but it was so much fun to see that you were actually on the right path the whole time
Cleverly done ✨
I like it!
genuinely feeling happy that I got to the same result
I did it a little differently: Since the 20-square is exactly 4 times the area of the 5-square, we can cut it up into four squares, each of area 5.
The side length of each 5-square is clearly √5. So the side length of the big square is 3√5.
The small white triangle is similar to the pink triangle, as Andy showed (right triangles with alternate interior angles).
So the ratio of the long leg to the short leg in both triangles must be the same.
In the white triangle, the long leg is 2√5 and the short leg is √5. So the ratio is 2:1.
The long leg of the pink triangle is s (the side of the pink square), so the short leg must be s/2.
The hypotenuse of the pink triangle is the side length of the big square which we already know is 3√5. So, by the pythagorean theorem, s² + (s/2)² = (3√5)² .
(3√5)² = 45, so the equation simplifes to 5s²/4 = 45. Solving for s², we get (4/5) x 45 = 36. Finished; no need to find s, itself.
I got this one! Not the numbers, but the principle. I got there through the way you prove that line theorem. Top corner blue (a)+top corner pink(b) = 90, they're both right triangles, so angle sum of 180, 90 for the right angle, so the 90 left goes to a and b
You are amazing❤
I love this one
I couldve done it if i remembered the congruence thing! This was a good one
This give my some nostalgia of 8th grade geometry classes
Nice problem love your stuff dude
Cool problem! I got the angles of the triangles using soh cah toa and the inverse sin
I loved math in school. I can still follow along. Thought we would have to get into trig on this one. Nice graphics on this one.
This is the first problem I solved on this channel, I've watch a few video and now I solved one. How exciting 😮
That is brilliant
im subbing to this amazing channe;. you would have been so helpful in 9th grade but its better late than never!
Very exciting!
Nice video. I solved the problem in a similar manner: Note that the side lengths of the three black squares are sqrt(5) times 1, 2 and 3.
Therefore the short catheters of all the similar triangles are half the size of the corresponding long catheters.
And the hypotenuses are sqrt(5) times the short catheters.
Then the short catheter of the pink triangle must be 3, and thus the long catheter is 6, and the pink square has area 36.
In my experience, cross multiplication is a technique that students consistently misapply so I avoid using it when teaching.
The first blue triangle has ratio 1:2:sqrt(5) and is congruent with the first pink triangle, of which the hypotenusa is 3sqrt(5). Hence the pink square has side 6, and thus an area of 36.
Wow! What a great math problem! Nice and exciting math videos by the way 🙂
Pretty much took those steps. One thing I noticed: In one of the last steps you cross-multiplied both sides, only to divide by one of the terms immediately after. Given that you only needed to multiply by one divisor to get the variable alone that "multiply-divide" step is unnecessary.
I like the prportion thing u did
That’s genius I love that
I solved it by using alpha = arctan(/sqrt(5) / /sqrt(20)), then s = cos(alpha) * (/sqrt(5)+/sqrt(20)) and then area = s^2
but your method was also cool
wow! i did see how to solve for the blue triangle but did not make the connection that the pink and blue triangles were similar :>
I constantly solve problems with a solution different from how the teacher solves them. This task is no exception.
I found that the side of the square (which is the hypotenuse of the triangle shaded in red) and it is equal to √45, the drawing clearly shows that the smaller catheter of the triangle shaded in red is 1/2 of the second
Let the small catheter be x, then the second one will be 2x.
So we get a triangle with x and 2x catheter whose hypotenuse is equal to √45.
According to the Pythagorean theorem, C^2=A^2+B^2, 45=x^2+4x^2 => 45=5x^2 | :5, => 9= x^2, x=3
(x is a small catheter, a larger catheter, which is also the side of the desired square 2 times larger, that is, the side is 6, the area is 36 sq units). The whole solution is based only on knowledge of the Pythagorean theorem.
I was thinking of using trigonometry to find the angles in the blue triangle and back again to get the relevant side of it's red counterpart... Much more elegant and efficient your way ^^
However for the length of the bigger square's size, I found calculating the total area and square rooting it to be faster more satisfying than the way you did it, even if it's not faster ^^
You easily find that the area is 45, meaning sides are sqrt(9*5), which is 3sqrt(5) ^^
This one feels conceptually pretty straightforward if you imagine a coordinate plane over the shapes. We know the slope of both bottom line segments on the square, and could calculate the intersection of those lines. But it does seem like some hairy algebra.
Sidelength of the given squares are √5 and √20 = 2√5. I called s = sidelength of the pink square and e = hypotenuse of right triangle on top of the 20-square. Then e = √[(2√5)² + (√5)²] = √[20 + 5] = 5. Due to similarity of triangles we have (2√5)/e = s/(3√5), hence s = (2√5)·(3√5)/5 = 6·5/5 = 6, therefore area = 6² = 36 square units.
Hooboy. I got the right answer, but got there by a bit different path. My path was a bit more convoluted, haha! This was fun to figure out though! 🤓
I solved it by setting a system of equations with what he had as the pink triangle netting the first equation:
a^2+(b+5)^2=3sqrt(5)
and a second triangle in the top left of the big box that determined the diagonal distance must be sqrt(10), so the second equation was:
a^2+b^2=10
not necessarily a better way but brings more information about the dimensions of the pink square once you solve the system of equations (either a=3 or -3 overlapping satisfies) :)
I like your funny words magic man
Cool✨
That excited me.
Another thrilling mathematic adventure!
great 👍👍
Exciting!
Your voice is incredible! Can you please record an audio book of Elements?
The happiest mathematician in the world:
Hey Andy where can I find some more of these type of questions? Really want to try some more
His voice is so great
Lovely how clean the solution ends up being, no roots or fractions at all.
Not having the time to do this myself right now, but I wonder if the identity works out to
square 1: x
square 2: x^2 - x
square 3: (x + 1)^2
?
So a 6, 30, would end up being 49? Or maybe just these numbers are special…
Work was slow and I did have time actually.
This does not work out :(
smaller squares area 3 and 6 make a larger square with area 6+4√2 instead of my presumed area of 16
Edit: I did find the cool extension of the pretty continuation though!
The sides of the large square work out to be (x-1)+√(x-1), so the “clean” solutions end up being when the smallest squares area is exactly 1 more than a perfect square,
so
10 and 90 == 144
17 and 272 == 400
26 and 650 == 900
…
How exciting.
You could just calculate the sides of the 5 and 20 square then the side of the white square and then you reverse Pythagoras to find out the right triangle sides which have also pink square side
In India, we learn algebra like this- when any operator on one side goes to the other side of the equation, it changes to the opposite operator. Like + to -, * to / and vice versa. I think it’s a bit faster that doing it on both sides.
pure love!
Does anyone know what is the program that he's using in the screen, I want to use to teach my students 😊😊 , & thanks in advance
I am an SSC aspirant and trust me these questions are good
How exciting
Nice one
It's so satisfying when you see your calculation gives whole numbers as the answer 🤌
there is an easier way to do this because radical 20 simplifies down to 2 radical 5 so you can add the radicals to get 3 radical five and square that for the area. please tell me if im wrong.
This was a fun one and I put a box around the final answer of 36. They are also 30-60-90 triangles.
HOW? EXCITING.
trully an exciting problem
Awesome!
You make it fun watching your explanations. Are you a math teacher?
I did it the same way, but misread the 5, 20 as side lengths rather than areas.
My method was almost identical, except I use an intermediate cosine out of laziness.
don't you need to subtract the area of the thick black line that divides the space?
I wish Andy was around when i was in Mu Alpha Theta, would've been way better than the Russian manuals we had to use 😂
I actually knew how to do this one!
Fun!
Nice.