# Characterization of a right triangle

Problem. Show that a triangle $ABC$ is a right triangle if and only if $\sin A + \sin B + \sin C = \cos A + \cos B + \cos C + 1$.

Vietnam 1981

Solution. (By removeablesingularity) $(\Rightarrow)$ Suppose that $ABC$ is a a right triangle. Then $\{A,B,C\}=\{\theta, 90^\circ - \theta, 90^\circ \}$. Check that $\sin A + \sin B + \sin C = \sin \theta + \cos \theta + 1$ $\cos A + \cos B + \cos C = \cos \theta + \sin \theta + 0$,

and hence the equation will be satisfied. $(\Leftarrow)$ Now, suppose that the equation is satisfied. If the left-hand-side (LHS) is squared, we obtain $\sin^2 A + \sin^2 B + \sin^2 C + 2 (\sin A \sin B + \sin B \sin C + \sin C \sin A)$

and if the right-hand-side (RHS) is squared, we obtain $= \cos^2 A + \cos^2 B + \cos^2 C + 2(\cos A \cos B + \cos B \cos C + \cos C \cos A) + 2\cos A + 2 \cos B + 2 \cos C + 1$.

Recall the trigonometric identities:

• $\cos^2 x - \sin^2 x = \cos (2x)$
• $\cos(x+y) = \cos x \cos y - \sin x \sin y$
• $\cos(180^\circ - \theta) + \cos \theta = 0$ (for instance, $\cos(A+B)+\cos C = 0$)

The equation above after squared then can rewritten as $\cos(2A) + \cos(2B) + \cos(2C)=-1$.

Let us write $C=180^\circ - A - B$. Then, $\cos(2A) +\cos(2B) + \cos(2A+2B) + 1 = 0$ $\cos(2A)+\cos(2B) +\cos(2A) \cos(2B) + 1= \sin(2A) \sin(2B)$ $(1+\cos(2A))(1+\cos(2B))=\sin(2A) \sin(2B)$ $2 \cos^2 A \cdot 2 \cos^2 B = 2 \sin A \cos A \cdot 2 \sin B \cos B$

which factorizes to $4 \cos A \cdot \cos B \cdot (\cos A \cos B - \sin A \sin B) = 0$ $4 \cos A \cdot \cos B \cdot \cos(A+B) = 0$ $4 \cos A \cdot \cos B \cdot \cos C = 0$

Hence, $\cos A=0$ or $\cos B = 0$ or $\cos C=0$ which consecutively implies $A=90^\circ$, $B = 90^\circ$, or $C=90^\circ$. Hence, if the equation is satisfied, triangle $ABC$ has to be a right triangle. $\blacksquare$ truth seeker

## 2 thoughts on “Characterization of a right triangle”

1. removablesingularity says:

Squaring both sides and a manipulation gives $cos 2A + cos 2B + cos 2C = -1$
writing $C = \pi - (A+B)$ yield $cos 2A + cos 2B + cos2A cos2B + 1 = sin 2A sin 2B$
or $(cos 2A + 1)(cos 2B + 1) = sin 2A sin 2B$.
This gives $4 cos^2 A cos^2 B = 4 sin A cos A sin B cos B$ or $cos A cos B cos(A+B) = 0$.

2. nivotko says:

Mantap Kak. Terima kasih hehe.