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:

$\textbf{(A)}\ \text{the center of the inscribed circle} \qquad \\ \textbf{(B)}\ \text{the center of the circumscribed circle}\qquad\\ \textbf{(C)}\ \text{such that the three angles formed at the point each be }{120^\circ}\qquad\\ \textbf{(D)}\ \text{the intersection of the altitudes of the triangle}\qquad\\ \textbf{(E)}\ \text{the intersection of the medians of the triangle}$

Solution

$[asy] import geometry; point A = (0,0); point B = (3,1); point C = (2,4); point P = (A+B+C)/3; draw(A--B--C--A); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NW); dot(P); label("P",P,E); draw(A--P); draw(B--P); draw(C--P); [/asy]$

$\fbox{\textbf{(E) }the intersection of the medians of the triangle}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1957 AHSME Problems/Problem 26

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