1955 AMC 12 Problem 29

Attempt Problem 29 of the 1955 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1955 AMC 12 solutions, or check the answer key.

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29.

In the figure PA\overline{PA} is tangent to semicircle SAR;SAR; PB\overline{PB} is tangent to semicircle RBT;RBT; SRTSRT is a straight line; the arcs are indicated in the figure. Angle APBAPB is measured by:

12(ab)\dfrac12(a-b)

12(a+b)\dfrac12(a+b)

(ca)(db)(c-a)-(d-b)

aba-b

a+ba+b

Answer: E
Concepts:tangent-tangent anglecircle arcsangle measure
Difficulty rating: 2310
Small Hint:

Draw PR\overline{PR}, which is tangent to both semicircles at RR

Big Hint:

Use the tangent-tangent angle theorem on each circle, then use a+c=b+d=180a+c=b+d=180^\circ

Solution:

The line PRPR is tangent to both semicircles at their common endpoint R.R. For the larger circle, the tangent-tangent angle theorem gives APR=180a=c. \angle APR=180^\circ-a=c. For the smaller circle, the same theorem gives RPB=180b=d. \angle RPB=180^\circ-b=d. Thus the reflex angle from PA\overline{PA} to PB\overline{PB} through PR\overline{PR} has measure c+d.c+d. Hence the other angle between the tangents is 360(c+d)=(180c)+(180d)=a+b, \begin{aligned} 360^\circ-(c+d) &=(180^\circ-c)\\ &\quad{}+(180^\circ-d)\\ &=a+b, \end{aligned} because each upper semicircle has measure 180.180^\circ.

Thus, the correct answer is E.

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