1954 AHSME Problems/Problem 19

Problem 19

If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:

$\textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles}\\ \textbf{(D)}\ \text{are always acute angles}\\ \textbf{(E)}\ \text{are always unequal to each other}$

Solution

Call the three angles in the triangle, $\theta$ , $\alpha$ and $180-\theta-\alpha$ Then there are three iscoscles triangles, and by angles chasing, we get that the three angles are $90-\frac{\alpha}{2}$ , $\frac{\theta+\alpha}{2}$ , and $90-\frac{\theta}{2}$ , which are all acute because, $\theta, \alpha>0$ , $\theta+\alpha<180\implies\frac{\theta+\alpha}{2}<90$

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1954 AHSME Problems/Problem 19

Problem 19

Solution