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

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==Solution==
 
==Solution==
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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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Let <math>CH</math> be the altitude of the triangle through vertex <math>C</math> and <math>GP</math> be the distance from <math>G</math> to <math>AB</math>, with the point <math>P</math> laying on <math>AB</math>.
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Using Thales' intercept theorem, we derive the proportion:
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<math>\frac{GP}{CH} = \frac{GM}{CM}</math>
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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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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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Answer: D
  
 
==See Also==
 
==See Also==

Latest revision as of 08:45, 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 (ProblemsAnswer KeyResources)
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
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