Difference between revisions of "1982 AHSME Problems/Problem 21"
MRENTHUSIASM (talk | contribs) (Recreated this page and wished to type up the solutions.) |
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draw(M--C--A--B--C^^B--N); | draw(M--C--A--B--C^^B--N); | ||
pair point=P; | pair point=P; | ||
| − | markscalefactor=0. | + | 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 | ||
\textbf{(B)}\ \frac 32s\sqrt2 \qquad | \textbf{(B)}\ \frac 32s\sqrt2 \qquad | ||
\textbf{(C)}\ 2s\sqrt2 \qquad | \textbf{(C)}\ 2s\sqrt2 \qquad | ||
| − | \textbf{(D)}\ \frac{ | + | \textbf{(D)}\ \frac{s\sqrt5}{2}\qquad |
| − | \textbf{(E)}\ \frac{ | + | \textbf{(E)}\ \frac{s\sqrt6}{2}</math> |
== Solution == | == Solution == | ||
| + | Suppose that <math>P</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{BN}.</math> Let <math>BN=x.</math> By the properties of centroids, we have <math>BP=\frac23 x.</math> | ||
| + | |||
| + | Note that <math>\triangle BPC\sim\triangle BCN</math> by AA. From the ratio of similitude <math>\frac{BP}{BC}=\frac{BC}{BN},</math> we get | ||
| + | <cmath>\begin{align*} | ||
| + | BP\cdot BN &= BC^2 \\ | ||
| + | \frac23 x\cdot x &= s^2 \\ | ||
| + | x^2 &= \frac32 s^2 \\ | ||
| + | x &= \boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}. | ||
| + | \end{align*}</cmath> | ||
| + | ~MRENTHUSIASM | ||
== See Also == | == See Also == | ||
{{AHSME box|year=1982|num-b=20|num-a=22}} | {{AHSME box|year=1982|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 22:42, 15 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
Suppose that 
is the intersection of 
and 
Let 
By the properties of centroids, we have 
Note that 
by AA. From the ratio of similitude 
we get
~MRENTHUSIASM
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. 
