Dynamic Geometry: P68
The diagram shows a yellow
-
-
right triangle. The blue and purple squares are inscribed in the triangle. Each square moves at such a rate so they meet at a unique point. Using the touching vertices of the square and the right angled vertex of the yellow triangle, we draw a green triangle. When the area of the green triangle is equal to
, the sum of the areas of the two squares can be expressed as
, where
and
are coprime positive integers. Find
the sum of digits
of
.
The answer is 8.
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1 solution
Label the diagram as follows, and place it on a Cartesian graph so that C is at the origin, A is at ( 4 , 0 ) , and B is at ( 0 , 3 ) . Also consider when the vertices of the squares G and K meet at the unique point P :
△ J L P , △ P H F , and △ F E A are all similar to △ B C A by AA similarity, which means P F P H = B A B C = 5 3 , so let P H = 3 k and P F = 5 k .
From the squares, J P = P H = 3 k and F E = P F = 5 k .
From similar triangles, L P = J P ⋅ B C A C = 3 k ⋅ 5 4 = 5 1 2 k = C D and E A = F E ⋅ B C C A = 5 k ⋅ 3 4 = 3 2 0 k .
Then C D + D E + E A = C A , or 5 1 2 k + 5 k + 3 2 0 k = 4 , which solves to k = 2 1 1 6 0 .
That means P has coordinates ( 5 1 2 k , 5 k ) = ( 5 1 2 ⋅ 2 1 1 6 0 , 5 ⋅ 2 1 1 6 0 ) = ( 2 1 1 1 4 4 , 2 1 1 3 0 0 ) .
Since B P = ( 2 1 1 1 4 4 − 0 , 2 1 1 3 0 0 − 3 ) = ( 2 1 1 1 4 4 , − 2 1 1 3 3 3 ) , K follows ( x , y ) = ( 0 , 3 ) + t ( 2 1 1 1 4 4 , − 2 1 1 3 3 3 ) , so K x = 2 1 1 1 4 4 t and K y = 3 − 2 1 1 3 3 3 t .
That means C K = ( K x − 0 , K y − 0 ) = ( 2 1 1 1 4 4 t , 3 − 2 1 1 3 3 3 t ) .
Since A P = ( 2 1 1 1 4 4 − 4 , 2 1 1 3 0 0 − 0 ) = ( − 2 1 1 7 0 0 , 2 1 1 3 0 0 ) , G follows ( x , y ) = ( 4 , 0 ) + t ( − 2 1 1 7 0 0 , 2 1 1 3 0 0 ) , so G x = 4 − 2 1 1 7 0 0 t and G y = 2 1 1 3 0 0 t .
That means C G = ( G x − 0 , G y − 0 ) = ( 4 − 2 1 1 7 0 0 t , 2 1 1 3 0 0 t ) .
The area of △ K C G is then A △ K C G = 2 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2 1 1 1 4 4 t 4 − 2 1 1 7 0 0 t 3 − 2 1 1 3 3 3 t 2 1 1 3 0 0 t ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 1 ∣ ∣ ∣ ∣ 2 1 1 1 4 4 t ⋅ 2 1 1 3 0 0 t − ( 3 − 2 1 1 3 3 3 t ) ( 4 − 2 1 1 7 0 0 t ) ∣ ∣ ∣ ∣ = 1 4 4 0 2 8 3 1 , which solves to t = 3 6 0 2 1 1 .
The side of the blue square is G y = 2 1 1 3 0 0 t = 2 1 1 3 0 0 ⋅ 3 6 0 2 1 1 = 6 5 , so its area is A blue = ( 6 5 ) 2 = 3 6 2 5 .
The side of the purple square is L K J K ⋅ K x = 4 5 ⋅ 2 1 1 1 4 4 t = 2 1 1 1 8 0 ⋅ 3 6 0 2 1 1 = 2 1 , so its area is A purple = ( 2 1 ) 2 = 4 1 .
The sum of the areas of the two squares is 3 6 2 5 + 4 1 = 1 8 1 7 , so p = 1 7 , q = 1 8 , q + p = 3 5 , and the sum of the digits of 3 5 is 3 + 5 = 8 .