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 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png