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 "1982 AHSME Problems/Problem 14"
(→‎Problem 14:)
Line 5: Line 5:
 
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle  <math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length
 
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle  <math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length
  
−
<math>[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("</math>A<math>", A, W); label("</math>B<math>", B, SE); label("</math>C<math>", C, NE); label("</math>D<math>", D, dir(0)); label("</math>P<math>", P, S); label("</math>N<math>", N, S); label("</math>O<math>", O, S); label("</math>E<math>", E, dir(120)); label("</math>F<math>", F, NE); label("</math>G<math>", G, dir(100));[/asy]</math>
+
[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("<math>A</math>", A, W); label("<math>B</math>", B, SE); label("<math>C</math>", C, NE); label("<math>D</math>", D, dir(0)); label("<math>P</math>", P, S); label("<math>N</math>", N, S); label("<math>O</math>", O, S); label("<math>E</math>", E, dir(120)); label("<math>F</math>", F, NE); label("<math>G</math>", G, dir(100));[/asy]
  
 
==Solution:==
 
==Solution:==

Revision as of 14:04, 22 August 2015

1982 AHSME Problems/Problem 14

Problem 14:

In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB, BC$, and $CD$ are diameters of circle $O, N$, and $P$, respectively. Circles $O, N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length

[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, dir(0)); label("$P$", P, S); label("$N$", N, S); label("$O$", O, S); label("$E$", E, dir(120)); label("$F$", F, NE); label("$G$", G, dir(100));[/asy]

Solution:

Since $GP$ is 15, $AP$ is 75, and $\angle{AGP}=90$, $AG=15\sqrt{24}$.

Now drop an altitude from $N$ to $AG$ at point $H$. $AN=45$, and since $\triangle{AGP}$ is similar to $\triangle{AHN}$. $NH=9$. $NE=NF=15$ so by Pythagorean Theorem, $EH=HF=12$. Thus $EF=\boxed{24}$. $\boxed{C}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_14&oldid=71732"