Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
{{solution}}
 
{{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>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>.
 +
 +
Using Thales' intercept theorem, we derive the proportion:
 +
 +
 +
<math>\frac{GP}{CH} = \frac{GM}{CM}</math>
 +
 +
 +
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> .
 +
 +
 +
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
  
 
==See Also==
 
==See Also==

Revision as of 15:10, 29 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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

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 in 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 in 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1962_AHSME_Problems/Problem_15&oldid=90051"