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 "1981 AHSME Problems/Problem 19"

(created solutions page (Accidentally put it in main page before) what the hell were the answer choices?)
 
(problem statement was missing explanation of theta.)
Line 1: Line 1:
 
==Problem 19==
 
==Problem 19==
−
In <math>\triangle ABC</math>, <math>M</math> is the midpoint of side <math>BC</math>, <math>AN</math> bisects <math>\angle BAC</math>, and <math>BN\perp AN</math>. If sides <math>AB</math> and <math>AC</math> have lengths <math>14</math> and <math>19</math>, respectively, then find <math>MN</math>.
+
In <math>\triangle ABC</math>, <math>M</math> is the midpoint of side <math>BC</math>, <math>AN</math> bisects <math>\angle BAC</math>, <math>BN\perp AN</math>, and <math>\theta</math> is the measure of <math>\angle BAC</math>. If sides <math>AB</math> and <math>AC</math> have lengths <math>14</math> and <math>19</math>, respectively, then find <math>MN</math>.
  
 
<asy>
 
<asy>

Revision as of 11:05, 28 June 2025

Problem 19

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$, and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then find $MN$.

[asy] size(150); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(30)); label(rotate(angle(dir(A--C)))*"$19$", A--C, dir(A--C)*dir(90)); label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90)); [/asy]

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ \dfrac{5}{2}\qquad\textbf{(C)}\ \dfrac{5}{2}-\sin\theta\qquad\textbf{(D)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right)$

Solution

Extend $BN$ to meet $AC$ at $Q$. Then $\triangle BNM \sim \triangle BQC$, so $BN=NQ$ and $QC=19-AQ=2MN$.

Since $\angle ANB=90^\circ = \angle ANQ$, $\angle BAN=\angle NAQ$ (since $AN$ is an angle bisector) and $\triangle ANB$ and $\triangle ANQ$ share side $AN$, $\triangle ANB \cong \triangle ANQ$. Thus $AQ=14$, and so $MN=\frac{19-AQ}{2}=\frac{5}{2}$, hence our answer is $\fbox{B}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_19&oldid=252600"