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

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 $\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.$

@@ 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>.
-Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, <math>\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 \\
 &= 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} \\

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

Revision as of 11:36, 1 August 2020