Difference between revisions of "1954 AHSME Problems/Problem 19"

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The same method applies to <math>\angle FED</math> and <math>\angle FDE</math>, which means <math>\triangle DEF</math> is acute - hence our answer is <math>\fbox{D}</math>.
 
The same method applies to <math>\angle FED</math> and <math>\angle FDE</math>, which means <math>\triangle DEF</math> is acute - hence our answer is <math>\fbox{D}</math>.
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==See Also==
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{{AHSME 50p box|year=1954|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 00:28, 28 February 2020

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
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All AHSME Problems and Solutions


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