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Difference between revisions of "1956 AHSME Problems/Problem 10"

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passes through the other two vertices. The side <math>AC</math> extended through <math>C</math> intersects the circle
 
passes through the other two vertices. The side <math>AC</math> extended through <math>C</math> intersects the circle
 
at <math>D</math>. The number of degrees of angle <math>ADB</math> is:
 
at <math>D</math>. The number of degrees of angle <math>ADB</math> is:
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<math>(A) 15 (B) 30 (C) 60 (D) 90 (E) 120</math>
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<math>\text{(A)} 15 \quad  \text{(B)}30 \quad \text{(C)}60 \quad \text{(D)}90 \quad \text{(E)}120</math>
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== Solution ==
 
== Solution ==
 
<asy>
 
<asy>

Latest revision as of 20:32, 12 February 2021

Problem

A circle of radius $10$ inches has its center at the vertex $C$ of an equilateral triangle $ABC$ and passes through the other two vertices. The side $AC$ extended through $C$ intersects the circle at $D$. The number of degrees of angle $ADB$ is:

$\text{(A)} 15 \quad \text{(B)}30 \quad \text{(C)}60 \quad \text{(D)}90 \quad \text{(E)}120$

Solution

[asy] import olympiad; draw(circle((0,0),10)); dot((0,0)); label("C", (1,-1)); dot((5,5sqrt(3))); dot((-5,-5sqrt(3))); draw((-5,-5sqrt(3))--(5,5sqrt(3))); label("A",(6,5sqrt(3)+1)); label("D",(-6,-5sqrt(3)-1)); draw((0,0)--(10,0)); label("10",(1.5,2.5sqrt(3)+1)); dot((10,0)); label("B",(11,-1)); draw((5,5sqrt(3))--(10,0)); draw(anglemark((10,0),(0,0),(5,5sqrt(3)),60)); label( "60°", (2.5,1.25)); draw((-5,-5sqrt(3))--(10,0)); draw(anglemark((10,0),(-5,-5sqrt(3)),(5,5sqrt(3)),60)); label("?",(-2.5,-5sqrt(3)+2.5)); [/asy]

$ABC$ is an equilateral triangle, so ∠$C$ must be $60$°. Since $D$ is on the circle and ∠$ADB$ contains arc $AB$, we know that ∠$D$ is $30$° $\implies \fbox{B}$.

See Also

1956 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
All AHSME Problems and Solutions

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