Suppose the sides of the yellow and blue triangles are a and b, then the sum of their areas is (a²√3)/4+(b²√3)/4=100, and the area of the green triangle is (ab*sin60)/2=(ab√3)/4, and the square of the length of the side of the pink triangle is a²+b²-2abcos60=a²+b²-ab, and the area is (a²+b²-ab)*√3/4, so the sum of the areas of the pink and green triangles is (ab√3)/4+(a²+b²-ab)*√3/4=(a²+b²)*√3)4=100
the answer is obviously 100 But that is because the two bottom triangles are equal the top two triangles must be equal to them too and the question implies that the answer is always the same. But, doing it properly... the area of an equilateral triangle is length squared times root3 all over 4 yellow length is a, area is A blue length is b, area is B purple length is d, area is D green area is C (a^2*root3)/4+(b^2*root3)/4=100 length d is the hypotenuse of a right angled triangle with the bottom length (a+b)/2 and the other side being height of b-height of a, height of an equilateral triangle is (length*root3)/2 so using pythag... d^2=((a+b)/2)^2+(((b-a)root3)/2)^2 which simplifies to d=root(a^2+b^2-ab) which makes the area D=((a^2+b^2-ab)*root3)/4 D=(a^2*root3)/4+(b^2*root3)/4-(ab*root3)/4 as per the above the first section 100, so D=100-(ab*root3)/4 the area of the green triangle is half of length a times length b times sin lowest angle equilateral triangles have 60 degree angles, the angle is 180-two of these angles, which is 60, so C=(absin60)/2 which goes to C=(ab*root3)/4 which makes C+D=100-(ab*root3)/4+(ab*root3)/4=100
The one at the end, hmm. I can empirically see that it should also be 100. I mean, if yellow = blue, then green and purple are congruent. If yellow = 0 and blue = 100, then green = 0 and purple = 100. No matter how you look at it, it's 100. I tried using law of cosines but that turned into a nightmare. What we learn from that is if the side of the yellow triangle is y and the side of the blue one is b, the side of the purple one is sqrt((y^2 + b^2)/(yb)). Not yet seeing how that's useful in a proof though.
First, love your content. Second, wondering if I could request a video on a puzzle after the agg-vent series. It is easy to prove that the angle bisectors of the equal angles of an isosceles triangle are equal - but can you prove given that the angle bisectors of a triangle are of equal length, that the triangle is isosceles?
Answer to the next question: Sin60° = (√3)/2 If the side lengths of the yellow and blue equilateral triangles are x and y (Sin60°)(x^2 + y^2)/2 = 100 (x^2 + y^2) = 400/(√3) Green area = (Sin60°)(xy)/2 = xy(√3)/4 Side length of the pink equilateral triangle = √[(x/2 + y/2)^2 + ((y/2)√3 - (x/2)√3)^2] = √[(1/4)(x^2 + y^2 + 2xy) + (3/4)(x^2 + y^2 - 2xy)] = √(x^2 + y^2 - xy) = √(400/(√3) - xy) Pink area = (Sin60°)(400/(√3) - xy)/2 = ((√3)/4)(400/(√3) - xy) = 100 - (√3)/4)xy Pink + green = 100
Tomorrow thw ans would be 100 I think We can find the sides of the two triangle which would be x and 2x and then we can get x^2 from there and then. When we can get the area of the green and pink in terms of x and then substitute the value of x^2 in there
I'm amazed at how many times you've had to "figure" out that the height of an equilateral triangle with an inscribed angle is 3r. It's just funny that this ends up being a part of most of these! Id suggest making a video showing this to be the case, but all of your aggvents would be 30 seconds long :D Also, not calling it liTTle-Arrrr is the real crime here.
Hi Andy, I wonder if you find this one fun: ua-cam.com/video/5Yaq7O3D9dY/v-deo.html The problem says, KN = 9 cm; ABC is an isosceles triangle (AB=BC; AC is the base); BD and CK are medians. They want us to find MD. Thanks!
That jump scared me😂
It's all about that tangency.
Andy is behind schedule: :(
Andy has to upload multiple times to catch up: :)
When I get bored, I watch Andy's channel.
Need to catch up.. It's the 21st and only at 16th puzzle.. l
Wdym? It's 22nd december
@@CuriousFragdifferent Timezones meaning this guy is in a timezone behind yours 👍
Ah, the VSauce open. Very clever.
Hey Vsauce! Andy Math here.
i was thinking the same...lol
Suppose the sides of the yellow and blue triangles are a and b, then the sum of their areas is (a²√3)/4+(b²√3)/4=100, and the area of the green triangle is (ab*sin60)/2=(ab√3)/4, and the square of the length of the side of the pink triangle is a²+b²-2abcos60=a²+b²-ab, and the area is (a²+b²-ab)*√3/4, so the sum of the areas of the pink and green triangles is (ab√3)/4+(a²+b²-ab)*√3/4=(a²+b²)*√3)4=100
the answer is obviously 100
But that is because the two bottom triangles are equal the top two triangles must be equal to them too and the question implies that the answer is always the same.
But, doing it properly... the area of an equilateral triangle is length squared times root3 all over 4
yellow length is a, area is A
blue length is b, area is B
purple length is d, area is D
green area is C
(a^2*root3)/4+(b^2*root3)/4=100
length d is the hypotenuse of a right angled triangle with the bottom length (a+b)/2 and the other side being height of b-height of a,
height of an equilateral triangle is (length*root3)/2 so using pythag...
d^2=((a+b)/2)^2+(((b-a)root3)/2)^2
which simplifies to
d=root(a^2+b^2-ab)
which makes the area
D=((a^2+b^2-ab)*root3)/4
D=(a^2*root3)/4+(b^2*root3)/4-(ab*root3)/4 as per the above the first section 100, so
D=100-(ab*root3)/4
the area of the green triangle is half of length a times length b times sin lowest angle
equilateral triangles have 60 degree angles, the angle is 180-two of these angles, which is 60, so
C=(absin60)/2
which goes to C=(ab*root3)/4
which makes C+D=100-(ab*root3)/4+(ab*root3)/4=100
how exciting
The one at the end, hmm. I can empirically see that it should also be 100. I mean, if yellow = blue, then green and purple are congruent. If yellow = 0 and blue = 100, then green = 0 and purple = 100. No matter how you look at it, it's 100. I tried using law of cosines but that turned into a nightmare. What we learn from that is if the side of the yellow triangle is y and the side of the blue one is b, the side of the purple one is sqrt((y^2 + b^2)/(yb)). Not yet seeing how that's useful in a proof though.
day17
Iet A=sqrt3/4
1.yellow+blue=A(a^2+b^2)
2.pink+green=A(c^2+ab)
3.c^2=(b·sqrt3/2)^2+(a-b/2)^2
=a^2+b^2-ab
yellow+blue=pink+green=100😊
the sum of pink and green area is 100
your math is always prettier than mine lol
First, love your content. Second, wondering if I could request a video on a puzzle after the agg-vent series. It is easy to prove that the angle bisectors of the equal angles of an isosceles triangle are equal - but can you prove given that the angle bisectors of a triangle are of equal length, that the triangle is isosceles?
Another beautiful solution! But you're falling behind on days?
Answer to the next question:
Sin60° = (√3)/2
If the side lengths of the yellow and blue equilateral triangles are x and y
(Sin60°)(x^2 + y^2)/2 = 100
(x^2 + y^2) = 400/(√3)
Green area = (Sin60°)(xy)/2 = xy(√3)/4
Side length of the pink equilateral triangle = √[(x/2 + y/2)^2 + ((y/2)√3 - (x/2)√3)^2] =
√[(1/4)(x^2 + y^2 + 2xy) + (3/4)(x^2 + y^2 - 2xy)] = √(x^2 + y^2 - xy) = √(400/(√3) - xy)
Pink area = (Sin60°)(400/(√3) - xy)/2 = ((√3)/4)(400/(√3) - xy) = 100 - (√3)/4)xy
Pink + green = 100
really liking the beard action on my man
How Aggciting!
Tomorrow thw ans would be 100 I think
We can find the sides of the two triangle which would be x and 2x and then we can get x^2 from there and then. When we can get the area of the green and pink in terms of x and then substitute the value of x^2 in there
0:45 X???? ITS NOT LITTLE R????
yay
I'm amazed at how many times you've had to "figure" out that the height of an equilateral triangle with an inscribed angle is 3r.
It's just funny that this ends up being a part of most of these! Id suggest making a video showing this to be the case, but all of your aggvents would be 30 seconds long :D
Also, not calling it liTTle-Arrrr is the real crime here.
He only says "little r" when "big R" also exists
Is anyone with 60√3 or Am I the only one. All the other people are answering 100 so anyone else with 60√3
Andy Math lore
where he has didn't post for 3 days btween day 14 and 15 because he was using Maths to save the world.
Can you do zebra puzzles it would be so fun to watch
What is a zebra puzzle?
First one i solved!
36? 9/4 pir2
Next problem: 100
I think tomorrow's answer is 100.
Too similar to one you just did.
36
Hi Andy, I wonder if you find this one fun: ua-cam.com/video/5Yaq7O3D9dY/v-deo.html
The problem says, KN = 9 cm; ABC is an isosceles triangle (AB=BC; AC is the base); BD and CK are medians. They want us to find MD.
Thanks!