1986 AHSME Problems/Problem 26
Problem
It is desired to construct a right triangle in the coordinate plane so that its legs are parallel
to the 
and 
axes and so that the medians to the midpoints of the legs lie on the lines 
and 
. The number of different constants 
for which such a triangle exists is
Noitulos
In any right triangle with legs parallel to the axes, one median to the midpoint of a leg has slope 
times that of the other. This can easily be shown with coordinates: any triangle of this sort may be labelled with right angle at 
, other vertices 
and 
, and thus midpoints 
and 
, so that the slopes are 
and 
, thus showing that one is times the other as required.
Thus in our problem, is either 
or 
. In fact, both are possible, and each for infinitely many triangles. We shall show this for 
, and the argument is analogous for 
. Take any right triangle with legs parallel to the axes and a hypotenuse with slope 
, e.g. the triangle with vertices 
, 
, and 
. Then quick calculations show that the medians to the legs have slopes 
and 
. Now translate the triangle (without rotating it) so that its medians intersect at the point where the lines 
and 
intersect. This forces the medians to lie on these lines (since their slopes are determined, and now we force them to go through a particular point; a slope and a point uniquely determine a line). Finally, for any central dilation of this triangle (a larger or smaller triangle with the same centroid and sides parallel to this one's sides), the medians will still lie on these lines, showing the "infinitely many" part of the result.
Hence, to sum up, can in fact be both or 
, which is exactly 
values, i.e. 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 27 | |
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