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Difference between revisions of "1957 AHSME Problems/Problem 26"

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<math>\fbox{\textbf{(E) }the intersection of the medians of the triangle}</math>.
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Suppose the triangle is <math>\triangle ABC</math> with the described point in its interior being <math>P</math>, as in the diagram. First, suppose that the smaller triangles have equal area (say <math>\tfrac{2A}3</math>, where <math>A=[\triangle ABC]</math>). Then, by rearranging the area formula for a triangle, we see that the distance from <math>P</math> to side <math>\overline{BC}</math> is <math>\tfrac{2A}{3BC}</math>. Similarly, we can see that the distance from <math>A</math> to <math>\overline{BC}</math> is <math>\tfrac{2A}{BC}</math>. Thus, <math>P</math> is <math>\tfrac1 3</math> of the distance that <math>A</math> is from <math>\overline{BC}</math>. We can use the same logic for points <math>B</math> and <math>C</math> 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 <math>P</math> is the intersection of the medians.
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Regarding the sufficiency of this condition, because the centroid is <math>\tfrac2 3</math> 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 <math>P</math> is the centroid.
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Thus, our answer is <math>\fbox{\textbf{(E) }the intersection of the medians of the triangle}</math>.
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To disprove the other answers, try to draw counterexamples in extreme cases (so that it is obvious that that option is incorrect).
  
 
== See Also ==
 
== See Also ==

Revision as of 15:54, 25 July 2024

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]

Suppose the triangle is $\triangle ABC$ with the described point in its interior being $P$, as in the diagram. First, suppose that the smaller triangles have equal area (say $\tfrac{2A}3$, where $A=[\triangle ABC]$). Then, by rearranging the area formula for a triangle, we see that the distance from $P$ to side $\overline{BC}$ is $\tfrac{2A}{3BC}$. Similarly, we can see that the distance from $A$ to $\overline{BC}$ is $\tfrac{2A}{BC}$. Thus, $P$ is $\tfrac1 3$ of the distance that $A$ is from $\overline{BC}$. We can use the same logic for points $B$ and $C$ 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 $P$ is the intersection of the medians.

Regarding the sufficiency of this condition, because the centroid is $\tfrac2 3$ 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 $P$ is the centroid.

Thus, our answer is $\fbox{\textbf{(E) }the intersection of the medians of the triangle}$.

To disprove the other answers, try to draw counterexamples in extreme cases (so that it is obvious that that option 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

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