Difference between revisions of "1982 AHSME Problems/Problem 21"
MRENTHUSIASM (talk | contribs) (Recreated this page and wished to type up the solutions.) |
MRENTHUSIASM (talk | contribs) (→Solution 1) |
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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 1 == |
+ | 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 | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>O</math> be the centroid. Knowing that there are two medians, we need to find the length of the third one. Therefore, we draw the median <math>AP</math> such that <math>O</math> is on <math>AP</math>. Then, it follows that <math>BP = CP = OP = \frac{s}{2}</math> by Thales's Theorem, and that <math>AO = s</math>. So, <math>AP = \frac{3}{2}s</math>, which gives us the idea that <math>CA = s\sqrt{2}</math>. | ||
+ | |||
+ | Since <math>N</math> is the median that cuts <math>CA</math>, we find out that <math>CN = AN = \frac{s\sqrt{2}}{2}</math>. Finally, using Pythagorean again gives <math>BN=\boxed{{\textbf{(E)}\ \frac{s\sqrt6}{2}}}</math>. | ||
+ | |||
+ | ~elpianista227 | ||
== 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 01:30, 9 December 2024
Contents
[hide]Problem
In the adjoining figure, the triangle is a right triangle with
. Median
is perpendicular to median
,
and side
. The length of
is
Solution 1
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
Solution 2
Let be the centroid. Knowing that there are two medians, we need to find the length of the third one. Therefore, we draw the median
such that
is on
. Then, it follows that
by Thales's Theorem, and that
. So,
, which gives us the idea that
.
Since is the median that cuts
, we find out that
. Finally, using Pythagorean again gives
.
~elpianista227
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.