# 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

For the sake of clarity, let the outermost triangle be $ABC$ with incircle tangency points $D$, $E$, and $F$ on $BC$, $AC$ and $AB$ respectively. Let $\angle A=\alpha$, and $\angle B=\beta$. Because $\triangle AFE$ and $\triangle BDF$ are isosceles, $\angle AFE=\frac{180-\alpha}{2}$ and $\angle BFD=\frac{180-\beta}{2}$. So $\angle DFE=180-\frac{180-\alpha}{2}-\frac{180-\beta}{2}=\frac{\alpha+\beta}{2}$, and since $\alpha + \beta <180^\circ$, $\angle DFE$ is acute.

The same method applies to $\angle FED$ and $\angle FDE$, which means $\triangle DEF$ is acute - hence our answer is $\fbox{D}$.

## See Also

 1954 AHSC (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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