1956 AHSME Problems/Problem 49

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,

$\angle AOB 180^\circ - \angle BAO - \angle ABO 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} \frac{\angle BAP + \angle ABP}{2}.$

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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1956 AHSME Problems/Problem 49