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

(problem statement was missing explanation of theta.)
 
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<asy>
 
<asy>
−
size(150);
+
size(200);
 
defaultpen(linewidth(0.7)+fontsize(10));
 
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);
 
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);
Line 13: Line 13:
 
label("$B$", B, dir(point--B));
 
label("$B$", B, dir(point--B));
 
label("$C$", C, dir(point--C));
 
label("$C$", C, dir(point--C));
−
label("$M$", M, dir(point--M));
+
label("$M$", M, S);
 
label("$N$", N, dir(30));
 
label("$N$", N, dir(30));
 
label(rotate(angle(dir(A--C)))*"$19$", A--C, dir(A--C)*dir(90));
 
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));
+
label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(270));
 
</asy>
 
</asy>
  
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== Solution ==
 
== Solution ==
−
Extend <math>BN</math> to meet <math>AC</math> at <math>Q</math>. Then <math>\triangle BNM \sim \triangle BQC</math>, so <math>BN=NQ</math> and <math>QC=19-AQ=2MN</math>.
 
  
−
Since <math>\angle ANB=90^\circ = \angle ANQ</math>, <math>\angle BAN=\angle NAQ</math> (since <math>AN</math> is an angle bisector) and <math>\triangle ANB</math> and <math>\triangle ANQ</math> share side <math>AN</math>, <math>\triangle ANB \cong \triangle ANQ</math>. Thus <math>AQ=14</math>, and so <math>MN=\frac{19-AQ}{2}=\frac{5}{2}</math>, hence our answer is <math>\fbox{B}</math>.
+
<asy>
 +
size(200);
 +
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(270));
 +
</asy>
 +
 
 +
Extend <math>BN</math> to meet <math>AC</math> at <math>Q</math>. Then <math>\triangle ANB \cong \triangle ANQ</math>. So <math>AQ=AB=14</math>, and <math>CQ = 19-14=5</math>. Also, <math>NB = NQ</math> so <math>N</math> is the midpoint of <math>\overline{BQ}</math>. Thus <math>\triangle BMN \sim \triangle BCQ</math>, and therefore <math>MN = \frac{1}{2}CQ = \boxed{(\textbf{B})\ \frac{5}{2}}</math>.
 +
 
 +
==See also==
 +
 
 +
{{AHSME box|year=1981|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Latest revision as of 17:41, 20 August 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(200); 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, S); 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(270)); [/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(200); 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(270)); [/asy]

Extend $BN$ to meet $AC$ at $Q$. Then $\triangle ANB \cong \triangle ANQ$. So $AQ=AB=14$, and $CQ = 19-14=5$. Also, $NB = NQ$ so $N$ is the midpoint of $\overline{BQ}$. Thus $\triangle BMN \sim \triangle BCQ$, and therefore $MN = \frac{1}{2}CQ = \boxed{(\textbf{B})\ \frac{5}{2}}$.

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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