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

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== Solution ==
 
== Solution ==
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Call the three angles in the triangle, <math>\theta</math>, <math>\alpha</math> and <math>180-\theta-\alpha</math>
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For the sake of clarity, let the outermost triangle be <math>ABC</math> with incircle tangency points <math>D</math>, <math>E</math>, and <math>F</math> on <math>BC</math>, <math>AC</math> and <math>AB</math> respectively. Let <math>\angle A=\alpha</math>, and <math>\angle B=\beta</math>. Because <math>\triangle AFE</math> and <math>\triangle BDF</math> are isosceles, <math>\angle AFE=\frac{180-\alpha}{2}</math> and <math>\angle BFD=\frac{180-\beta}{2}</math>. So <math>\angle DFE=180-\frac{180-\alpha}{2}-\frac{180-\beta}{2}=\frac{\alpha+\beta}{2}</math>, and since <math>\alpha + \beta <180^\circ</math>, <math>\angle DFE</math> is acute.
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Then there are three iscoscles triangles, and by angles chasing, we get that the three angles are <math>90-\frac{\alpha}{2}</math>, <math>\frac{\theta+\alpha}{2}</math>, and <math>90-\frac{\theta}{2}</math>, which are all acute because, <math>\theta, \alpha>0</math>, <math>\theta+\alpha<180\implies\frac{\theta+\alpha}{2}<90</math>
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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>.
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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 (Problems • Answer Key • Resources)
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


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