Difference between revisions of "1981 AHSME Problems/Problem 19"

Latest revision as of 03:00, 9 July 2018

Problem 19

In $\triangle ABC$ , $M$ is the midpoint of side $BC$ , $AN$ bisects $\angle BAC$ , and $BN\perp AN$ . 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 $\fbox{B}$ .

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Difference between revisions of "1981 AHSME Problems/Problem 19"

Latest revision as of 03:00, 9 July 2018

Problem 19

Solution