A B C D is a convex quadrilateral such that :
A D A E = A B B F = B C G C = C D H D = 3 1
If A B C D area is equal to 1 and E F G H area is equal to b a with co-prime a , b find a + b
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Let be a square of side lenght 1 and area 1. According to the statement, we can draw a new square which has the same length as the hypothenuse of the right triangle of side lenght 1/3 and 2/3.
Thus, this length measures: sqrt ((1/3) ^ 2 + (2/3) ^ 2).
Now, by calculating the area of this new square, we find 5/9. Plus 5 + 9 = 14.
The area of the outside quadrilateral is 2 1 × A B × A D × sin A + 2 1 × C B × C D × sin C = 2 1 × B A × B C × sin B + 2 1 × D A × D C × sin D = 1 △ A E F = 2 1 × A E × A F × sin A = 2 1 × 3 1 × A D × 3 2 × A B = 9 2 × 2 1 × A D × A B △ C G H = 2 1 × C G × C H × sin C = 9 2 × 2 1 × C B × C D × sin C △ A E F + △ C G H = 9 2 × ( 2 1 × A B × A D × sin A + 2 1 × C B × C D × sin C ) = 9 2 × 1 △ B F G + △ D E H = 9 2 So the inside quadrilateral is 1 − 9 2 − 9 2 = 9 5
Instead of finding directly E F G H area i'll find the triangle areas and subtract from A B C D
Consider the diagonal B D we know that A B D + B C D = 1 If you pay attention A B E have the same height as A B D but it's relative base is 3 A D so its area is equal to 3 A B D Now, look at A E F and A E B :
A E F has the same height as A E B but it's base is equal to 2 × 3 A B ∴
A E F area is equal to A E B × 3 2 that is equal to A B D × 9 2 )
If you do the same with triangle B C D you will get the same pattern. The sum of the "opposite" triangles it's equal to A B D × 9 2 + B C D × 9 2 = 9 2 ( A B D + B C D ) = 9 2 ( 1 ) Skipping unecessary calculations we get that A F E + B F G + C G H + D H E = 9 4 so E F G H = 1 − 9 4 = 9 5
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Let a r e a ( Δ A C D ) = A , a r e a ( Δ A C B ) = B , a r e a ( Δ A B D ) = C , a r e a ( Δ B C D ) = D
Now, a r e a ( Δ A C D ) a r e a ( Δ E C D ) = A D E D = 3 2 ⟹ a r e a ( Δ E C D ) = 3 2 a r e a ( Δ A C D ) = 3 2 A
Also, a r e a ( Δ E C D ) a r e a ( Δ E H D ) = D C D H = 3 1 ⟹ a r e a ( Δ E H D ) = 3 1 a r e a ( Δ E C D ) = 3 1 ⋅ 3 2 A = 9 2 A ⋯ E q . 1
In the similar manner, a r e a ( Δ B F G ) = 9 2 B ⋯ E q . 2 a r e a ( Δ A E F ) = 9 2 C ⋯ E q . 3 a r e a ( Δ C G H ) = 9 2 D ⋯ E q . 4
Adding all these 4 equations we get,
a r e a ( Δ E H D ) + a r e a ( Δ B F G ) + a r e a ( Δ A E F ) + a r e a ( Δ C G H ) = 9 2 ( A + B ) + 9 2 ( C + D ) ⟹ a r e a ( A B C D ) − a r e a ( E F G H ) = 9 2 ⋅ 1 + 9 2 ⋅ 1 = 9 4 ⟹ a r e a ( E F G H ) = a r e a ( A B C D ) − 9 4 = 1 − 9 4 = 9 5
So, a = 5 , b = 9 ⟹ a + b = 1 4