1957 AHSME Problems/Problem 1

The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:

$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 3$

Solution

$[asy] size(2cm); draw((-3,0)--(0,4)--(3,0)--cycle); draw((0,0)--(0,4), red); draw((-3,0)--(0.84, 2.88), green); draw((-3,0)--(1.5, 2), green); draw((-3,0)--(1.636, 1.818), green); draw((3,0)--(-0.84, 2.88), blue); draw((3,0)--(-1.5, 2), blue); draw((3,0)--(-1.636, 1.818), blue); [/asy]$

As shown in the diagram above, all nine altitudes, medians, and interior angle bisectors are distinct, except for the three coinciding lines from the vertex opposite to the base. Thusly, there are $7$ distinct lines, so our answer is $\boxed{\textbf{(B)}}$ , and we are done.

1957 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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.

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1957 AHSME Problems/Problem 1

Solution

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