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Difference between revisions of "1962 AHSME Problems/Problem 15"

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==Solution==
 
==Solution==
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{{solution}}
 
  
 
Let <math>CM</math> be the median through vertex <math>C</math>, and let <math>G</math> be the point of intersection of the triangle's medians.  
 
Let <math>CM</math> be the median through vertex <math>C</math>, and let <math>G</math> be the point of intersection of the triangle's medians.  
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The fraction <math>\frac{GM}{CM}</math> in any triangle is
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The fraction <math>\frac{GM}{CM}</math> in any triangle is equal to <math>\frac{1}{3}</math> . Therefore <math>GP = \frac{CH}{3}</math> .
  
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equal to <math>\frac{1}{3}</math> . Therefore <math>GP = \frac{CH}{3}</math> .
 
  
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Since the problem states that the vertex <math>C</math> is moving on a straight line, the length of <math>CH</math> is a constant value. That means that the length of <math>GP</math> is also a constant. Therefore the point <math>G</math> is moving on a straight line.
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Since the problem states that the vertex <math>C</math> is moving in a straight line, the length of <math>CH</math> is a constant value. That means that the length of <math>GP</math> is also a constant. Therefore the point <math>G</math> is moving in a straight line.
 
  
 
Answer: D
 
Answer: D

Latest revision as of 07:45, 30 January 2018

Problem

Given triangle $ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line, the intersection point of the three medians moves on:

$\textbf{(A)}\ \text{a circle}\qquad\textbf{(B)}\ \text{a parabola}\qquad\textbf{(C)}\ \text{an ellipse}\qquad\textbf{(D)}\ \text{a straight line}\qquad\textbf{(E)}\ \text{a curve here not listed}$

Solution

Let $CM$ be the median through vertex $C$, and let $G$ be the point of intersection of the triangle's medians.

Let $CH$ be the altitude of the triangle through vertex $C$ and $GP$ be the distance from $G$ to $AB$, with the point $P$ laying on $AB$.

Using Thales' intercept theorem, we derive the proportion:


$\frac{GP}{CH} = \frac{GM}{CM}$


The fraction $\frac{GM}{CM}$ in any triangle is equal to $\frac{1}{3}$ . Therefore $GP = \frac{CH}{3}$ .


Since the problem states that the vertex $C$ is moving on a straight line, the length of $CH$ is a constant value. That means that the length of $GP$ is also a constant. Therefore the point $G$ is moving on a straight line.

Answer: D

See Also

1962 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
All AHSME Problems and Solutions

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