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 "1986 AHSME Problems/Problem 29"

(→‎Solution)
(→‎Solution)
Line 11: Line 11:
  
 
==Solution==
 
==Solution==
 +
Assume we have a scalene triangle <math>ABC</math>. Arbitrarily, let 12 be the height to base <math>AB</math> and 4 be the height to base <math>AC</math>. Due to area equivalences, the base <math>AC</math> must be three times the length of <math>AB</math>.
 +
 +
Let the base <math>AB</math> be <math>x</math>, thus making <math>AC = 3x</math>. Thus, setting the final height to base <math>BC</math> to <math>h</math>, we note that <math>(BC * h)/2 = (3x * 4)/2 = 6x</math>. Thus, <math>h = (12x / BC)</math>. We note that to maximize <math>h</math> we must minimize <math>BC</math>. Using the triangle inequality, <math>BC + AB > AC</math>, thus <math>BC + x > 3x</math> or <math>BC > 2x</math>. The minimum value of <math>BC</math> is <math>2x</math>, which would output <math>h = 6</math>. However, because <math>BC</math> must be larger than <math>2x</math>, the minimum integer height must be <math>5</math>.
 +
 
<math>\fbox{B}</math>
 
<math>\fbox{B}</math>
  

Revision as of 19:51, 19 September 2015

Problem

Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12$. If the length of the third altitude is also an integer, what is the biggest it can be?

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{none of these}$

Solution

Assume we have a scalene triangle $ABC$. Arbitrarily, let 12 be the height to base $AB$ and 4 be the height to base $AC$. Due to area equivalences, the base $AC$ must be three times the length of $AB$.

Let the base $AB$ be $x$, thus making $AC = 3x$. Thus, setting the final height to base $BC$ to $h$, we note that $(BC * h)/2 = (3x * 4)/2 = 6x$. Thus, $h = (12x / BC)$. We note that to maximize $h$ we must minimize $BC$. Using the triangle inequality, $BC + AB > AC$, thus $BC + x > 3x$ or $BC > 2x$. The minimum value of $BC$ is $2x$, which would output $h = 6$. However, because $BC$ must be larger than $2x$, the minimum integer height must be $5$.

$\fbox{B}$

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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

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=1986_AHSME_Problems/Problem_29&oldid=72134"