In right , is the altitude to hypotenuse with and .
If in right the perimeter and the value of can be expressed as
, where and are coprime positive integers, find .
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For convenience let j = 2 5 (This is how I constructed the problem).
Note: j > = 2 ensures that △ A B C is a right triangle.
In △ A B C the altitude is the mean proportion between the segments of the hypotenuse
⟹ c 2 = x ( j c − x ) = ( j c ) x − x 2 ⟹ x 2 − ( j c ) x + c 2 = 0 ⟹
x = ( 2 j ± j 2 − 4 ) c
Using x = ( 2 j ± j 2 − 4 ) c ⟹ ∣ D C ∣ = ( 2 j − j 2 − 4 ) c and ∣ B D ∣ = ( 2 j + j 2 − 4 ) c
⟹ ∣ A C ∣ = ∣ A D ∣ 2 + ∣ D C ∣ 2 = 2 j 2 − j j 2 − 4 c
and
∣ A B ∣ = ∣ B D ∣ 2 + ∣ A D ∣ 2 = 2 j 2 + j j 2 − 4 c
Note: Using x = ( 2 j − j 2 − 4 ) c the values of ∣ A B ∣ and ∣ A C ∣ are just switched, so that ∣ A C ∣ = 2 j 2 + j j 2 − 4 and ∣ A B ∣ = 2 j 2 − j j 2 − 4 c .
Using j = 2 5 The Perimeter P = ( 2 3 5 + 5 ) c = c 2 ⟹
c ( c − ( 2 3 5 + 5 ) ) = 0 and c = 0 ⟹ c = 2 3 5 + 5 =
f d ∗ e + e ⟹ d + e + f = 1 0 .