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 "1990 AHSME Problems/Problem 14"

(Created page with "== Problem == An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at po...")
 
(→‎Solution)
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
<asy>
 +
draw(circle((0,0),1),black);
 +
draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot);
 +
draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot);
 +
draw(arc((0,1),.25,230,310));
 +
MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S);
 +
MP("x",(0,.8),S);
 +
</asy>
  
−
An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math>
+
An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC}=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math>
  
 
<math>\text{(A) } \frac{3\pi}{7}\quad
 
<math>\text{(A) } \frac{3\pi}{7}\quad
Line 10: Line 18:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{A}</math>
+
We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{540}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{540}{7}</math> degrees, the answer is <math>\dfrac{3\pi}{7}</math> which is <math>\fbox{A}.</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 13:09, 2 April 2023

Problem

[asy] draw(circle((0,0),1),black); draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot); draw(arc((0,1),.25,230,310)); MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S); MP("x",(0,.8),S); [/asy]

An acute isosceles triangle, $ABC$, is inscribed in a circle. Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$. If $\angle{ABC}=\angle{ACB}=2\angle{D}$ and $x$ is the radian measure of $\angle{A}$, then $x=$

$\text{(A) } \frac{3\pi}{7}\quad \text{(B) } \frac{4\pi}{9}\quad \text{(C) } \frac{5\pi}{11}\quad \text{(D) } \frac{6\pi}{13}\quad \text{(E) } \frac{7\pi}{15}$

Solution

We can make two equations (assume angle D is y): $y+2x=180$ and $4y+x=180$. We find that $x=\dfrac{540}{7}$. Now we have to convert this to radians. 360 degrees is $2\pi$ radians, so since we have $\dfrac{540}{7}$ degrees, the answer is $\dfrac{3\pi}{7}$ which is $\fbox{A}.$

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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=1990_AHSME_Problems/Problem_14&oldid=191711"