3 squares + 2 triangles

Geometry Level 3

The figure shows congruent three squares aligned with the sides of an equilateral triangle. The inner sides of the squares form another equilateral triangle. If the ratio of the area of the small triangle to the area of the large triangle is t t , submit 1 0 6 t \lfloor 10^6 t\rfloor .


The answer is 50180.

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2 solutions

Chew-Seong Cheong
Dec 20, 2020

Let the side length of the small equilateral triangle and square be 1 1 . Joint the top vertices of the two equilateral triangles with a line. Then we note that the side length of the large equilateral triangle is 1 + 2 3 1+2\sqrt 3 . Since the area of of a same shaped figure is directly promotional to the square of linear dimension, the ratio of the area of the small equilateral triangle to the area of the large equilateral triangle is:

t = Δ s m a l l Δ l a r g e = 1 2 ( 1 + 2 3 ) 2 = 1 13 + 4 3 0.050180139 1 0 6 t = 50180 t = \frac {\Delta_{\rm small}}{\Delta_{\rm large}} = \frac {1^2}{(1+2\sqrt 3)^2} = \dfrac 1{13+4\sqrt 3} \approx 0.050180139 \implies \lfloor 10^6t\rfloor = \boxed{50180}

It should be (13+4*sqrt(3)) in the denominator.

Correct it!

Vijay Simha - 5 months, 3 weeks ago

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Thanks. I have corrected it. Please try not to use so much "!" in English. It means that you are shouting at me to "Correct it!" like a teach to a small child.

Chew-Seong Cheong - 5 months, 3 weeks ago

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A fair complaint. Vijay's demand was made rudely.

Richard Desper - 5 months, 3 weeks ago

Sorry Chew.

Vijay Simha - 5 months, 2 weeks ago

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@Vijay Simha It is alright. I was not complaining but just to let you know. It is not sensitive in ohter languages such as Chinese but not English.

Chew-Seong Cheong - 5 months, 2 weeks ago
Sathvik Acharya
Dec 20, 2020

Let length of each side of X Y Z \triangle XYZ be equal to a a . Since the three squares share a common side with X Y Z \triangle XYZ , the length of each side the squares is also a a .

Notice that Z E = Z F ZE=ZF Z E B = Z F B = 9 0 \angle ZEB=ZFB=90^\circ B E Z B F Z \implies \triangle BEZ\cong \triangle BFZ So we have Z B E = Z B F = B 2 = 3 0 \angle ZBE=\angle ZBF=\frac{\angle B}{2}=30^\circ In B F Z \triangle BFZ , tan 3 0 = Z F B F = a B F \tan 30^\circ=\frac{ZF}{BF}=\frac{a}{BF} B F = a 3 \implies BF=a\sqrt{3} The length of each side of A B C \triangle ABC is A B = A G + G F + B F = a 3 + a + a 3 = a ( 2 3 + 1 ) AB=AG+GF+BF=a\sqrt{3}+a+a\sqrt{3}=a(2\sqrt{3}+1) Therefore, t = [ X Y Z ] [ A B C ] = 3 4 X Y 2 3 4 A B 2 = ( a a ( 2 3 + 1 ) ) 2 = 1 ( 2 3 + 1 ) 2 t=\frac{[\triangle XYZ]}{[\triangle ABC]}=\frac{\frac{\sqrt{3}}{4}\cdot XY^2}{\frac{\sqrt{3}}{4}\cdot AB^2}=\left ( \frac{a}{a(2\sqrt{3}+1)} \right)^2=\frac{1}{(2\sqrt3+1)^2} t 0.05018013 1 0 6 t = 50180 \implies t\approx 0.05018013 \implies \boxed{\lfloor 10^6\cdot t\rfloor = 50180}

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