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

(Created page with "== Problem 19== If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle: <math> \textbf{(A)}\ \text{are always e...")
 
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== Solution ==
 
== Solution ==
 
Call the three angles in the triangle, <math>\theta</math>, <math>\alpha</math> and <math>180-\theta-\alpha</math>
 
Call the three angles in the triangle, <math>\theta</math>, <math>\alpha</math> and <math>180-\theta-\alpha</math>
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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>
+
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>, so <math>\fbox{D}</math>

Revision as of 14:35, 6 June 2016

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, $\theta, \alpha>0$, $\theta+\alpha<180\implies\frac{\theta+\alpha}{2}<90$, so $\fbox{D}$

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