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Difference between revisions of "1986 AHSME Problems/Problem 26"

(Created page with "==Problem== It is desired to construct a right triangle in the coordinate plane so that its legs are parallel to the <math>x</math> and <math>y</math> axes and so that the medi...")
 
(Added a solution with explanation)
 
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==Solution==
 
==Solution==
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In ''any'' right triangle with legs parallel to the axes, one median to the midpoint of a leg has slope <math>4</math> times that of the other. This can easily be shown with coordinates: any triangle of this sort may be labelled with right angle at <math>P(a,b)</math>, other vertices <math>Q(a,b+2c)</math> and <math>R(a-2d,b)</math>, and thus midpoints <math>(a,b+c)</math> and <math>(a-d,b)</math>, so that the slopes are <math>\frac{c}{2d}</math> and <math>\frac{2c}{d} = 4(\frac{c}{2d})</math>, thus showing that one is <math>4</math> times the other as required.
  
 +
Thus in our problem, <math>m</math> is either <math>3 \times 4 = 12</math> or <math>3 \div 4 = \frac{3}{4}</math>. In fact, both are possible, and each for infinitely many triangles. We shall show this for <math>m=12</math>, and the argument is analogous for <math>m=\frac{3}{4}</math>. Take any right triangle with legs parallel to the axes and a hypotenuse with slope <math>12 \div 2 = 6</math>, e.g. the triangle with vertices <math>(0,0)</math>, <math>(1,0)</math>, and <math>(1,6)</math>. Then quick calculations show that the medians to the legs have slopes <math>12</math> and <math>3</math>. Now translate the triangle (without rotating it) so that its medians intersect at the point where the lines <math>y=12x+2</math> and <math>y=3x+1</math> intersect. This forces the medians to lie on these lines (since their slopes are determined, and now we force them to go through a particular point; a slope and a point uniquely determine a line). Finally, for any central dilation of this triangle (a larger or smaller triangle with the same centroid and sides parallel to this one's sides), the medians will still lie on these lines, showing the "infinitely many" part of the result.
 +
 +
Hence, to sum up, <math>m</math> can in fact be both <math>12</math> or <math>\frac{3}{4}</math>, which is exactly <math>2</math> values, i.e. <math>\boxed{C}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:05, 1 April 2018

Problem

It is desired to construct a right triangle in the coordinate plane so that its legs are parallel to the $x$ and $y$ axes and so that the medians to the midpoints of the legs lie on the lines $y = 3x + 1$ and $y = mx + 2$. The number of different constants $m$ for which such a triangle exists is

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ \text{more than 3}$


Solution

In any right triangle with legs parallel to the axes, one median to the midpoint of a leg has slope $4$ times that of the other. This can easily be shown with coordinates: any triangle of this sort may be labelled with right angle at $P(a,b)$, other vertices $Q(a,b+2c)$ and $R(a-2d,b)$, and thus midpoints $(a,b+c)$ and $(a-d,b)$, so that the slopes are $\frac{c}{2d}$ and $\frac{2c}{d} = 4(\frac{c}{2d})$, thus showing that one is $4$ times the other as required.

Thus in our problem, $m$ is either $3 \times 4 = 12$ or $3 \div 4 = \frac{3}{4}$. In fact, both are possible, and each for infinitely many triangles. We shall show this for $m=12$, and the argument is analogous for $m=\frac{3}{4}$. Take any right triangle with legs parallel to the axes and a hypotenuse with slope $12 \div 2 = 6$, e.g. the triangle with vertices $(0,0)$, $(1,0)$, and $(1,6)$. Then quick calculations show that the medians to the legs have slopes $12$ and $3$. Now translate the triangle (without rotating it) so that its medians intersect at the point where the lines $y=12x+2$ and $y=3x+1$ intersect. This forces the medians to lie on these lines (since their slopes are determined, and now we force them to go through a particular point; a slope and a point uniquely determine a line). Finally, for any central dilation of this triangle (a larger or smaller triangle with the same centroid and sides parallel to this one's sides), the medians will still lie on these lines, showing the "infinitely many" part of the result.

Hence, to sum up, $m$ can in fact be both $12$ or $\frac{3}{4}$, which is exactly $2$ values, i.e. $\boxed{C}$.

See also

1986 AHSME (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
All AHSME Problems and Solutions

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