1956 AHSME Problems/Problem 49: Difference between revisions
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Solution | ==Problem== | ||
Triangle <math>PAB</math> is formed by three tangents to circle <math>O</math> and <math>\angle APB = 40^{\circ}</math>; then <math>\angle AOB</math> equals: | |||
<math>\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}</math> | |||
==Solution== | |||
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>. | 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>. | ||
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Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, | Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, | ||
<math>\angle AOB | <math>\angle AOB = 180^\circ - \angle BAO - \angle ABO = | ||
180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} | 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} = | ||
180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} | 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}= | ||
\frac{\angle BAP + \angle ABP}{2}.</math> | \frac{\angle BAP + \angle ABP}{2}.</math> | ||
Finally, from triangle <math>ABP</math>, <math>\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ</math>, so <cmath>\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.</cmath> | Finally, from triangle <math>ABP</math>, <math>\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ</math>, so <cmath>\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.</cmath> | ||
== See Also == | |||
{{AHSME 50p box|year=1956|num-b=48|num-a=50}} | |||
[[Category:Introductory Algebra Problems]] | |||
{{MAA Notice}} | |||
Latest revision as of 00:32, 3 January 2024
Problem
Triangle 
is formed by three tangents to circle 
and 
; then 
equals:
Solution
First, from triangle 
, 
. Note that 
bisects 
(to see this, draw radii from to 
and 
creating two congruent right triangles), so 
. Similarly, 
.
Also, 
, and 
. Hence,
Finally, from triangle 
, 
, so ![\[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.\]](http://latex.artofproblemsolving.com/c/e/e/cee158004b042523864b03663557cb88324d27c8.png)
See Also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 48 |
Followed by Problem 50 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
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