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Difference between revisions of "1956 AHSME Problems/Problem 49"

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Solution 1
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==Problem==
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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:
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<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>
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==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,  
  
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<math>\angle AOB &= 180^\circ - \angle BAO - \angle ABO \\
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<math>\angle AOB = 180^\circ - \angle BAO - \angle ABO =
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&= 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} \\
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180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} =
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&= 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} \\
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180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}=
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&= \frac{\angle BAP + \angle ABP}{2}.</math>
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\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>
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== See Also ==
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{{AHSME 50p box|year=1956|num-b=48|num-a=50}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 00:32, 3 January 2024

Problem

Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40^{\circ}$; then $\angle AOB$ equals:

$\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}$


Solution

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 \[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.\]

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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