Square in Square 2

Geometry Level 3

In the diagram, A B C D ABCD is a square with an area of 25 25 , J K L M JKLM is a square with an area of 18 18 , and A I = B E = C F = D H AI = BE = CF = DH .

Find A E B E \dfrac{AE}{BE} .

(Diagram not drawn to scale.)


The answer is 6.

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3 solutions

David Vreken
Aug 17, 2019

Since A B C D ABCD has an area of 25 25 , A B = 5 AB = 5 , and since J K L M JKLM has an area of 18 18 , J M = 3 2 JM = 3\sqrt{2} . Let x = B E = A I x = BE = AI and y = A M = B J y = AM = BJ .

Since A M B \triangle AMB is a right triangle, by Pythagorean's Theorem y 2 + ( 3 2 + y ) 2 = 5 2 y^2 + (3\sqrt{2} + y)^2 = 5^2 which solves to y = 2 2 y = \frac{\sqrt{2}}{2} , and since H A B A M B \triangle HAB \sim \triangle AMB by the AA similarity postulate, x 5 = y 3 2 + y \frac{x}{5} = \frac{y}{3\sqrt{2} + y} so that with y = 2 2 y = \frac{\sqrt{2}}{2} this solves to x = 5 7 x = \frac{5}{7} .

Therefore the ratio A E B E = 5 5 7 5 7 = 6 \frac{AE}{BE} = \frac{5 - \frac{5}{7}}{\frac{5}{7}} = \boxed{6} .

Chew-Seong Cheong
Aug 16, 2019

We notice that the green and pink triangles are congruent. Therefore the area of one of these triangles is A = [ A B C D ] [ J K L M ] 4 = 25 18 4 = 7 4 A_\triangle = \dfrac {[ABCD]-[JKLM]}4 = \dfrac {25-18}4 = \dfrac 74 . Let A M = a AM = a and B M = b BM=b ; then A = 1 2 a b = 7 4 a b = 7 2 A_\triangle = \frac 12 ab = \frac 74 \implies ab = \frac 72 . Since the area of square A B C D ABCD is 25 25 , A B = 5 \implies AB = 5 . By Pythagorean theorem we have:

a 2 + b 2 = 25 Since a b = 7 2 a 2 + ( 7 2 a ) 2 = 25 Multiply both sides by 4 a 2 4 a 4 + 49 = 100 a 2 Rearrange 4 a 4 100 a 2 + 49 = 0 ( 2 a 2 1 ) ( 2 a 2 49 ) = 0 \begin{aligned} a^2 + b^2 & = 25 & \small \color{#3D99F6} \text{Since }ab = \frac 72 \\ a^2 + \left(\frac 7{2a}\right)^2 & = 25 & \small \color{#3D99F6} \text{Multiply both sides by }4a^2 \\ 4a^4 + 49 & = 100a^2 & \small \color{#3D99F6} \text{Rearrange} \\ 4a^4 - 100a^2 + 49 & = 0 \\ (2a^2-1)(2a^2-49) & = 0 \end{aligned}

Note that both a a and b b satisfy the equation. { a = 1 2 b = 7 2 \implies \begin{cases} a = \frac 1{\sqrt 2} \\ b = \frac 7{\sqrt 2} \end{cases} , since b > a b > a . Now we note that A M B \triangle AMB and E J B \triangle EJB are similar. Therefore,

B E B J = A B B M B E b J M = 5 b Since [ J K L M ] = 18 , J M = 3 2 B E = 5 ( 7 2 3 2 ) 7 2 = 5 7 \begin{aligned} \frac {BE}{BJ} & = \frac {AB}{BM} \\ \frac {BE}{b-JM} & = \frac 5b & \small \color{#3D99F6} \text{Since }[JKLM] = 18, \implies JM = 3\sqrt 2 \\ \implies BE & = \frac {5\left(\frac 7{\sqrt 2}-3\sqrt 2\right)}{\frac 7{\sqrt 2}} = \frac 57 \end{aligned}

Therefore, A E B E = A B B E B E = 5 5 7 1 = 6 \dfrac {AE}{BE} = \dfrac {AB-BE}{BE} = \dfrac 5{\frac 57} - 1 = \boxed 6 .

Nice approach!

David Vreken - 1 year, 9 months ago

ah \text{ah} , df \text{df} and bi \text{bi} are two letter variable names.

ah = y = 5 5 x δ \text{ah}=y=5-\frac{5 x}{\delta }

df = y = δ x 5 \text{df}=y=\frac{\delta x}{5}

bi = y = ( 5 δ ) + δ x 5 \text{bi}=y=(5-\delta )+\frac{\delta x}{5}

m = { x , y } /. Solve [ ah bi ] [ [ 1 ] ] { 5 δ 2 δ 2 + 25 , 5 ( δ 2 5 δ + 25 ) δ 2 + 25 } m=\{x,y\}\text{/.}\, \text{Solve}[\text{ah}\land \text{bi}][[1]] \Rightarrow \left\{\frac{5 \delta ^2}{\delta ^2+25},\frac{5 \left(\delta ^2-5 \delta +25\right)}{\delta ^2+25}\right\}

l = { x , y } /. Solve [ ah df ] [ [ 1 ] ] { 25 δ δ 2 + 25 , 5 δ 2 δ 2 + 25 } l=\{x,y\}\text{/.}\, \text{Solve}[\text{ah}\land \text{df}][[1]] \Rightarrow \left\{\frac{25 \delta }{\delta ^2+25},\frac{5 \delta ^2}{\delta ^2+25}\right\}

δ = ( δ /. Solve [ SquaredEuclideanDistance [ m , l ] = 18 ] ) [ [ 1 ] ] 5 7 \delta =(\delta \text{/.}\, \text{Solve}[\text{SquaredEuclideanDistance}[m,l]=18])[[1]] \Rightarrow \frac57

5 δ δ 6 \frac{5-\delta }{\delta }\Rightarrow 6

Of course, the simplier answer is s ( 2 s 2 + s ) s 2 1 -\frac{s \left(\sqrt{2-s^2}+s\right)}{s^2-1} with s s being set to 18 25 \sqrt{\frac{18}{25}} giving 6 6 .

With central square areas of 1, 5, 10 and 18, rational results of 1 3 \frac13 , 1 1 , 2 2 and 6 6 are availed, respectively.

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