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

Author: nivotko

truth seeker

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