Difference between revisions of "1989 AHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | 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 | + | 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}} | {{MAA Notice}} |
Latest revision as of 17:04, 21 October 2018
Problem
In , , , , is an altitude, and is a median. Then
Solution
We are told that is a median, so . Drop an altitude from to , adding point , and you can see that and are similar, implying , implying that and are congruent, so .
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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