Triangular Paper Folding

Geometry Level 3

The figure shown above is a piece paper in the shape of an equilateral triangle A B C ABC . Now the paper is folded along the crease E F EF with the vertex B B meeting the edge A C AC at point D D such that A D : D C = 3 : 2 AD:DC=3:2 . Find B E : B F BE:BF .

The answer can be expressed in the form of a b \cfrac{a}{b} , where a a and b b are positive coprime integers. Compute a + b a+b .

This is part of the set Fun With Problem-Solving .


The answer is 15.

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2 solutions

David Vreken
May 26, 2018

Since A D : D C AD:DC = 3 : 2 3:2 , let A D = 3 k AD = 3k and D C = 2 k DC = 2k , and each side of the equilateral triangle 5 k 5k . Also, let B E = x BE = x and B F = y BF = y . Then E D = x ED = x and F D = y FD = y , and since each side of the equilateral triangle is 5 k 5k , A E = 5 k x AE = 5k - x and C F = 5 k y CF = 5k - y . Also, since A B C \triangle ABC is an equilateral triangle, D C F = 60 ° \angle DCF = 60° and D A E = 60 ° \angle DAE = 60° .

By law of cosines on A D E \triangle ADE , D E 2 = A D 2 + A E 2 2 A D A E cos E A D DE^2 = AD^2 + AE^2 - 2 \cdot AD \cdot AE \cdot \cos \angle EAD , or x 2 = ( 3 k ) 2 + ( 5 k x ) 2 2 3 k ( 5 k x ) cos 60 ° x^2 = (3k)^2 + (5k - x)^2 - 2 \cdot 3k \cdot (5k - x) \cdot \cos 60° , which solves to x = 19 k 7 x = \frac{19k}{7} .

By law of cosines on C D F \triangle CDF , D F 2 = C D 2 + C F 2 2 C D C F cos D C F DF^2 = CD^2 + CF^2 - 2 \cdot CD \cdot CF \cdot \cos \angle DCF , or y 2 = ( 2 k ) 2 + ( 5 k y ) 2 2 2 k ( 5 k x ) cos 60 ° y^2 = (2k)^2 + (5k - y)^2 - 2 \cdot 2k \cdot (5k - x) \cdot \cos 60° , which solves to x = 19 k 8 x = \frac{19k}{8} .

Therefore, B E : B F = x y = 19 k 8 19 k 7 = 8 7 BE:BF = \frac{x}{y} = \frac{\frac{19k}{8}}{\frac{19k}{7}} = \frac{8}{7} , which means a = 8 a = 8 and b = 7 b = 7 , and a + b = 8 + 7 = 15 a + b = 8 + 7 = \boxed{15} .

Well done. Did not thought that cosine rule can be applied here too.

donglin loo - 3 years ago
Donglin Loo
May 24, 2018

To make things easier, we could let the side length of Δ A B C \Delta ABC be 5 5 .(The final answer demands ratio and the ratio will always be a constant value regardless of side length of triangles.). Then, we have A D = 3 , D C = 2 AD=3,DC=2 .

As shown above, Δ C F D Δ A D E \Delta CFD \sim \Delta ADE

A D C F = A E C D = D E D F \therefore \cfrac{AD}{CF}=\cfrac{AE}{CD}=\cfrac{DE}{DF}

3 5 m = 5 n 2 = n m \ \cfrac{3}{5-m}=\cfrac{5-n}{2}=\cfrac{n}{m}

5 n 2 = n m \cfrac{5-n}{2}=\cfrac{n}{m}

m = 2 n 5 n \Rightarrow m=\cfrac{2n}{5-n}

3 5 m = 5 n 2 \ \cfrac{3}{5-m}=\cfrac{5-n}{2}

3 5 2 n 5 n = 5 n 2 \ \cfrac{3}{5-\cfrac{2n}{5-n}}=\cfrac{5-n}{2}

5 ( 5 n ) 2 n = 6 \Rightarrow 5(5-n)-2n=6

25 7 n = 6 25-7n=6

n = 19 7 n=\cfrac{19}{7}

m = 2 19 7 5 19 7 = 2 19 35 19 = 19 8 \therefore m=\cfrac{2 \cdot \cfrac{19}{7}}{5-\cfrac{19}{7}}=\cfrac{2 \cdot19}{35-19}=\cfrac{19}{8}

Hence, B E B F = D E D F = 19 7 19 8 = 8 7 \cfrac{BE}{BF}=\cfrac{DE} {DF}=\cfrac{\cfrac{19}{7}}{\cfrac{19}{8}}=\cfrac{8}{7}

a + b = 8 + 7 = 15 a+b=8+7=15

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