The figure shown above is a piece paper in the shape of an equilateral triangle A B C . Now the paper is folded along the crease E F with the vertex B meeting the edge A C at point D such that A D : D C = 3 : 2 . Find B E : B F .
The answer can be expressed in the form of b a , where a and b are positive coprime integers. Compute a + b .
This is part of the set Fun With Problem-Solving .
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Well done. Did not thought that cosine rule can be applied here too.
To make things easier, we could let the side length of Δ A B C be 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 .
As shown above, Δ C F D ∼ Δ A D E
∴ C F A D = C D A E = D F D E
5 − m 3 = 2 5 − n = m n
2 5 − n = m n
⇒ m = 5 − n 2 n
5 − m 3 = 2 5 − n
5 − 5 − n 2 n 3 = 2 5 − n
⇒ 5 ( 5 − n ) − 2 n = 6
2 5 − 7 n = 6
n = 7 1 9
∴ m = 5 − 7 1 9 2 ⋅ 7 1 9 = 3 5 − 1 9 2 ⋅ 1 9 = 8 1 9
Hence, B F B E = D F D E = 8 1 9 7 1 9 = 7 8
a + b = 8 + 7 = 1 5
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Since A D : D C = 3 : 2 , let A D = 3 k and D C = 2 k , and each side of the equilateral triangle 5 k . Also, let B E = x and B F = y . Then E D = x and F D = y , and since each side of the equilateral triangle is 5 k , A E = 5 k − x and C F = 5 k − y . Also, since △ A B C is an equilateral triangle, ∠ D C F = 6 0 ° and ∠ D A E = 6 0 ° .
By law of cosines on △ A D E , D E 2 = A D 2 + A E 2 − 2 ⋅ A D ⋅ A E ⋅ cos ∠ E A D , or x 2 = ( 3 k ) 2 + ( 5 k − x ) 2 − 2 ⋅ 3 k ⋅ ( 5 k − x ) ⋅ cos 6 0 ° , which solves to x = 7 1 9 k .
By law of cosines on △ C D F , D F 2 = C D 2 + C F 2 − 2 ⋅ C D ⋅ C F ⋅ cos ∠ D C F , or y 2 = ( 2 k ) 2 + ( 5 k − y ) 2 − 2 ⋅ 2 k ⋅ ( 5 k − x ) ⋅ cos 6 0 ° , which solves to x = 8 1 9 k .
Therefore, B E : B F = y x = 7 1 9 k 8 1 9 k = 7 8 , which means a = 8 and b = 7 , and a + b = 8 + 7 = 1 5 .