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Difference between revisions of "1956 AHSME Problems/Problem 49"

(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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Line 3: Line 3:
 
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>.
  
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Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, \begin{align*}
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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 $ABO$, $\angle AOB = 180^\circ - \angle BAO - \angle ABO$. Note that $AO$ bisects $\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\angle BAO = \angle BAT/2$. Similarly, $\angle ABO = \angle ABR/2$.

Also, $\angle BAT = 180^\circ - \angle BAP$, and $\angle ABR = 180^\circ - \angle ABP$. 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 $ABP$, $\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ$, so \[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.\]

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