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Difference between revisions of "1989 AHSME Problems/Problem 7"

(Created page with "== Problem == In <math> \triangle ABC</math>, <math>\angle A = 100^\circ</math>, <math>\angle B = 50^\circ</math>, <math>\angle C = 30^\circ</math>, <math>\overline{AH}</math> i...")
 
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==Solution==
 
==Solution==
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We are told that <math>\overline{BM}</math> is a median, so <math>\overline{AM}=\overline{MC}</math>. Drop an altitude from <math>M</math> to <math>\overline{HC}</math>, adding point <math>N</math>, and you can see that <math>\triangle ACH</math> and <math>\triangle MCN</math> are similar, implying <math>\overline{HN}=\overline{NC}</math>, implying that <math>\triangle MNH</math> and <math>\triangle MNC</math> are similar, so <math>\angle MHC=\angle C=30^\circ</math>.
+
We are told that <math>\overline{BM}</math> is a median, so <math>\overline{AM}=\overline{MC}</math>. Drop an altitude from <math>M</math> to <math>\overline{HC}</math>, adding point <math>N</math>, and you can see that <math>\triangle ACH</math> and <math>\triangle MCN</math> are similar, implying <math>\overline{HN}=\overline{NC}</math>, implying that <math>\triangle MNH</math> and <math>\triangle MNC</math> are congruent, so <math>\angle MHC=\angle C=30^\circ</math>.
  
 
<asy>
 
<asy>
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label("M", (11,3), NE);
 
label("M", (11,3), NE);
 
label("N", (11,0), S);</asy>
 
label("N", (11,0), S);</asy>
 +
 +
== See also ==
 +
{{AHSME box|year=1989|num-b=6|num-a=8}} 
 +
 +
[[Category: Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:04, 21 October 2018

Problem

In $\triangle ABC$, $\angle A = 100^\circ$, $\angle B = 50^\circ$, $\angle C = 30^\circ$, $\overline{AH}$ is an altitude, and $\overline{BM}$ is a median. Then $\angle MHC=$

[asy] draw((0,0)--(16,0)--(6,6)--cycle); draw((6,6)--(6,0)--(11,3)--(0,0)); dot((6,6)); dot((0,0)); dot((11,3)); dot((6,0)); dot((16,0)); label("A", (6,6), N); label("B", (0,0), W); label("C", (16,0), E); label("H", (6,0), S); label("M", (11,3), NE);[/asy]

$\textrm{(A)}\ 15^\circ\qquad\textrm{(B)}\ 22.5^\circ\qquad\textrm{(C)}\ 30^\circ\qquad\textrm{(D)}\ 40^\circ\qquad\textrm{(E)}\ 45^\circ$

Solution

We are told that $\overline{BM}$ is a median, so $\overline{AM}=\overline{MC}$. Drop an altitude from $M$ to $\overline{HC}$, adding point $N$, and you can see that $\triangle ACH$ and $\triangle MCN$ are similar, implying $\overline{HN}=\overline{NC}$, implying that $\triangle MNH$ and $\triangle MNC$ are congruent, so $\angle MHC=\angle C=30^\circ$.

[asy] draw((0,0)--(16,0)--(6,6)--cycle); draw((6,6)--(6,0)--(11,3)--(0,0)); draw((11,3)--(11,0)); dot((6,6)); dot((0,0)); dot((11,3)); dot((6,0)); dot((16,0)); dot((11,0)); label("A", (6,6), N); label("B", (0,0), W); label("C", (16,0), E); label("H", (6,0), S); label("M", (11,3), NE); label("N", (11,0), S);[/asy]

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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