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Geometry Level 3

Let ∠ A \angle A and ∠ B \angle B be acute angles.

If sin ⁡ 2 A + sin ⁡ 2 B = sin ⁡ ( A + B ) = k \sin^2A + \sin^2B = \sin(A+B) = k , find k k


The answer is 1.

Alan Yan by Alan Yan , 20, USA

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1 solution

Alan Yan
Aug 28, 2015

Notice that if A and B are the acute angles of a right triangle, it always works. Therefore, if we can prove that if the given is true, then A + B = π 2 A + B = \frac{\pi}{2} , then we will be done with an answer of 1.

In order to prove this, let x = sin ⁡ A x = \sin A and y = sin ⁡ B y = \sin B .

sin ⁡ ( A + B ) = sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A = x 1 − y 2 + y 1 − x 2 \sin (A+B) = \sin A \cos B + \sin B \cos A = x\sqrt{1-y^2} + y\sqrt{1 - x^2}

We will prove that x 2 + y 2 = 1 x^2 + y^2 = 1 by contradiction

If x 2 + y 2 > 1 x^2 + y^2 > 1 , x 2 + y 2 = x 1 − y 2 + y 1 − x 2 < x x 2 + y 2 − y 2 + y x 2 + y 2 − x 2 x^2 + y^2 = x\sqrt{1-y^2} + y\sqrt{1 - x^2} < x\sqrt{x^2+y^2-y^2} + y\sqrt{x^2 + y^2 - x^2}

= x 2 + y 2 ⟹ x 2 + y 2 < x 2 + y 2 = x^2 + y^2 \implies x^2 + y^2 < x^2 + y^2 which is a contradiction.

If x 2 + y 2 < 1 x^2+y^2 < 1 , x 2 + y 2 = x 1 − y 2 + y 1 − x 2 > x x 2 + y 2 − y 2 + y x 2 + y 2 − x 2 x^2 + y^2 = x\sqrt{1-y^2} + y\sqrt{1 - x^2} > x\sqrt{x^2+y^2-y^2} + y\sqrt{x^2 + y^2 - x^2}

= x 2 + y 2 ⟹ x 2 + y 2 > x 2 + y 2 = x^2 + y^2 \implies x^2 + y^2 > x^2 + y^2 which is a contradiction.

Therefore, we know that x 2 + y 2 = 1 x^2 + y^2 = \boxed{1} and we have proven that A and B must be the acute angles of a right triangle.

(huricane)

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