1956 AMC 12 Problem 49

Question

Triangle

PAB

is formed by three tangents to circle

O

and

\angle APB=40^\circ;

then angle

AOB

equals:

45^\circ

50^\circ

55^\circ

60^\circ

70^\circ

The circle is tangent to one side of triangle

PAB

and the extensions of the other two, so

O

is the excenter opposite

P

The external angle bisectors at

A

and

B

form an angle of

90^\circ-\frac12\angle P

The circle lies opposite

P

across side

AB,

so

O

is the excenter opposite

P.

Thus

AO

and

BO

bisect the exterior angles at

A

and

B.

If the interior angles at

A

and

B

are

\alpha

and

\beta,

then triangle

AOB

has angles

90^\circ-\frac{\alpha}{2}

and

90^\circ-\frac{\beta}{2}

at

A

and

B.

Hence

\begin{aligned} \angle AOB &=180^\circ-\left(90^\circ-\frac\alpha2\right)\\ &\mathrel{\phantom{=}}-\left(90^\circ-\frac\beta2\right)\\ &=\frac{\alpha+\beta}{2} =\frac{180^\circ-40^\circ}{2}\\ &=70^\circ. \end{aligned}

Thus, the correct answer is E.

Problem 49 in other years