Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Problem)
Line 4: Line 4:
 
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:
 
<math>(A) 15 (B) 30 (C) 60 (D) 90 (E) 120</math>
 
<math>(A) 15 (B) 30 (C) 60 (D) 90 (E) 120</math>
 +
== Solution ==
 
<asy>
 
<asy>
 
import olympiad;
 
import olympiad;
Line 27: Line 28:
  
 
<math>ABC</math> is an equilateral triangle, so ∠<math>C</math> must be <math>60</math>°. Since <math>D</math> is on the circle and ∠<math>ADB</math> contains arc <math>AB</math>, we know that ∠<math>D</math> is <math>30</math>° <math>\implies \fbox{B}</math>.
 
<math>ABC</math> is an equilateral triangle, so ∠<math>C</math> must be <math>60</math>°. Since <math>D</math> is on the circle and ∠<math>ADB</math> contains arc <math>AB</math>, we know that ∠<math>D</math> is <math>30</math>° <math>\implies \fbox{B}</math>.
 +
 +
==See Also==
 +
 +
{{AHSME box|year=1956|num-b=9|num-a=11}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 20:30, 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: $(A) 15 (B) 30 (C) 60 (D) 90 (E) 120$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1956_AHSME_Problems/Problem_10&oldid=146267"