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, $ $\begin{aligned} ∠ A O B & = 180^{\circ} - ∠ B A O - ∠ A B O \\ = 180^{\circ} - \frac{∠ B A T}{2} - \frac{∠ A B R}{2} \\ = 180^{\circ} - \frac{180^{\circ} - ∠ B A P}{2} - \frac{180^{\circ} - ∠ A B P}{2} \\ = \frac{∠ B A P + ∠ A B P}{2} . \end{aligned}$ $ (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 $\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.$

Page

Toolbox

Search

1956 AHSME Problems/Problem 49