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1981 AHSME Problems/Problem 19

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

[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), Q=2*N-B; draw(N--B--A--N--M--C--A^^B--M); draw(N--Q); markscalefactor=0.1; draw(rightanglemark(B,N,A)); markscalefactor=0.3; draw(anglemark(B,A,N)); markscalefactor=0.2; draw(anglemark(N,A,Q)); markscalefactor=0.1; 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(-45)); label("$N$", N, dir(45)); label("$Q$", Q, dir(45)); label(rotate(angle(dir(A--Q)))*"$14$", A--Q, dir(A--Q)*dir(90)); label(rotate(angle(dir(Q--C)))*"$5$", Q--C, dir(Q--C)*dir(90)); label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90)); [/asy]

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 $\fbox{B}$.

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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

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