1956 AHSME Problems/Problem 49: Difference between revisions
Created page with "Solution 1 First, from triangle <math>ABO</math>, <math>\angle AOB = 180^\circ - \angle BAO - \angle ABO</math>. Note that <math>AO</math> bisects <math>\angle BAT</math> (to..." |
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First, from triangle <math>ABO</math>, <math>\angle AOB = 180^\circ - \angle BAO - \angle ABO</math>. Note that <math>AO</math> bisects <math>\angle BAT</math> (to see this, draw radii from <math>O</math> to <math>AB</math> and <math>AT,</math> creating two congruent right triangles), so <math>\angle BAO = \angle BAT/2</math>. Similarly, <math>\angle ABO = \angle ABR/2</math>. | First, from triangle <math>ABO</math>, <math>\angle AOB = 180^\circ - \angle BAO - \angle ABO</math>. Note that <math>AO</math> bisects <math>\angle BAT</math> (to see this, draw radii from <math>O</math> to <math>AB</math> and <math>AT,</math> creating two congruent right triangles), so <math>\angle BAO = \angle BAT/2</math>. Similarly, <math>\angle ABO = \angle ABR/2</math>. | ||
Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, \begin{align*} | Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, <math>\begin{align*} | ||
\angle AOB &= 180^\circ - \angle BAO - \angle ABO \\ | \angle AOB &= 180^\circ - \angle BAO - \angle ABO \\ | ||
&= 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} \\ | &= 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} \\ | ||
&= 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} \\ | &= 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} \\ | ||
&= \frac{\angle BAP + \angle ABP}{2}. | &= \frac{\angle BAP + \angle ABP}{2}. | ||
\end{align*} | \end{align*}</math> | ||
Finally, from triangle <math>ABP</math>, <math>\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ</math>, so <cmath>\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.</cmath> | Finally, from triangle <math>ABP</math>, <math>\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ</math>, so <cmath>\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.</cmath> | ||
Revision as of 11:36, 1 August 2020
Solution 1
First, from triangle 
, 
. Note that 
bisects 
(to see this, draw radii from 
to 
and 
creating two congruent right triangles), so 
. Similarly, 
.
Also, 
, and 
. Hence, $\begin{align*}
\angle AOB &= 180^\circ - \angle BAO - \angle ABO \\
&= 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} \\
&= 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} \\
&= \frac{\angle BAP + \angle ABP}{2}.
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Finally, from triangle 
, 
, so ![\[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.\]](http://latex.artofproblemsolving.com/c/e/e/cee158004b042523864b03663557cb88324d27c8.png)
