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Difference between revisions of "1982 AHSME Problems/Problem 21"

(Recreated this page and wished to type up the solutions.)
 
m
Line 9: Line 9:
 
draw(M--C--A--B--C^^B--N);
 
draw(M--C--A--B--C^^B--N);
 
pair point=P;
 
pair point=P;
−
markscalefactor=0.005;
+
markscalefactor=0.01;
 +
draw(rightanglemark(B,C,N));
 
draw(rightanglemark(C,P,B));
 
draw(rightanglemark(C,P,B));
 
label("$A$", A, dir(point--A));
 
label("$A$", A, dir(point--A));
Line 16: Line 17:
 
label("$M$", M, S);
 
label("$M$", M, S);
 
label("$N$", N, dir(C--A)*dir(90));
 
label("$N$", N, dir(C--A)*dir(90));
−
label("$s$", B--C, NW);</asy>
+
label("$s$", B--C, NW);
 +
</asy>
  
 
<math>\textbf{(A)}\ s\sqrt 2 \qquad  
 
<math>\textbf{(A)}\ s\sqrt 2 \qquad  

Revision as of 21:53, 13 September 2021

Problem

In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ$. Median $CM$ is perpendicular to median $BN$, and side $BC=s$. The length of $BN$ is

[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10));real r=54.72; pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M); draw(M--C--A--B--C^^B--N); pair point=P; markscalefactor=0.01; draw(rightanglemark(B,C,N)); draw(rightanglemark(C,P,B)); 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(C--A)*dir(90)); label("$s$", B--C, NW); [/asy]

$\textbf{(A)}\ s\sqrt 2 \qquad \textbf{(B)}\ \frac 32s\sqrt2 \qquad \textbf{(C)}\ 2s\sqrt2 \qquad \textbf{(D)}\ \frac{1}{2}s\sqrt5\qquad \textbf{(E)}\ \frac{1}{2}s\sqrt6$

Solution

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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