Difference between revisions of "1982 AHSME Problems/Problem 21"
MRENTHUSIASM (talk | contribs) m |
MRENTHUSIASM (talk | contribs) (→Problem) |
||
| Line 24: | Line 24: | ||
\textbf{(C)}\ 2s\sqrt2 \qquad | \textbf{(C)}\ 2s\sqrt2 \qquad | ||
\textbf{(D)}\ \frac{1}{2}s\sqrt5\qquad | \textbf{(D)}\ \frac{1}{2}s\sqrt5\qquad | ||
| − | \textbf{(E)}\ \frac{1}{2}s\sqrt6</math> | + | \textbf{(E)}\ \frac{1}{2}s\sqrt6</math> |
== Solution == | == Solution == | ||
Revision as of 22:00, 13 September 2021
Problem
In the adjoining figure, the triangle 
is a right triangle with 
. Median 
is perpendicular to median 
,
and side 
. The length of 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]](http://latex.artofproblemsolving.com/9/8/f/98f891d054bcb9cbfc2bc384c989c5b6725d2e09.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
