1957 AHSME Problems/Problem 26
Problem
From a point within a triangle, line segments are drawn to the vertices. A necessary and sufficient condition that the three triangles thus formed have equal areas is that the point be:
Solution
Suppose the triangle is with the described point in its interior being , as in the diagram. First, suppose that the smaller triangles have equal area (say , where ). Then, by rearranging the area formula for a triangle, we see that the distance from to side is . Similarly, we can see that the distance from to is . Thus, is of the distance that is from . We can use the same logic for points and and their respective opposite sides. The only point which is posiitioned in this way for all three points of the triangle is the centroid, so it is necessary that be the intersection of the medians.
Regarding the sufficiency of this condition, because the centroid is of the way along each of the medians of the triangle, the three smaller triangles, with the same base and a third of the height of the large triangle, have one third of the area of the larger triangle. Thus, they all have equal areas, so it is sufficient that point is the centroid.
Thus, our answer is .
To disprove the other answers, try to draw counterexamples in extreme cases (so that it is obvious that a specific answer choice is incorrect).
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.