Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1989 AHSME Problems/Problem 7: Difference between revisions

← Older edit
Elpianista227 (talk | contribs)
editor
40 edits
No edit summary
 
Line 39: Line 39:
label("N", (11,0), S);</asy>
label("N", (11,0), S);</asy>


==Solution 2==
Using Thales's Theorem, which states that the median drawn from the hypotenuse is half the hypotenuse, gives <math>MH = AM = AC</math>. Thus, <math>\triangle MHC</math> is isosceles. This gives <math>\angle MHC = \angle MCH = 30^\circ</math>.
== See also ==
== See also ==
{{AHSME box|year=1989|num-b=6|num-a=8}}   
{{AHSME box|year=1989|num-b=6|num-a=8}}   

Latest revision as of 08:34, 10 March 2025

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]

Solution 2

Using Thales's Theorem, which states that the median drawn from the hypotenuse is half the hypotenuse, gives $MH = AM = AC$. Thus, $\triangle MHC$ is isosceles. This gives $\angle MHC = \angle MCH = 30^\circ$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_7&oldid=244625"