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 "1954 AHSME Problems/Problem 35"
 
Line 1: Line 1:
 +
== Problem 35 ==
 +
 +
In the right triangle shown the sum of the distances <math>BM</math> and <math>MA</math> is equal to the sum of the distances <math>BC</math> and <math>CA</math>.
 +
If <math>MB = x, CB = h</math>, and <math>CA = d</math>, then <math>x</math> equals:
 +
 +
<asy>
 +
defaultpen(linewidth(.8pt)+fontsize(10pt));
 +
dotfactor=4;
 +
draw((0,0)--(8,0)--(0,5)--cycle);
 +
label("C",(0,0),SW);
 +
label("A",(8,0),SE);
 +
label("M",(0,5),N);
 +
dot((0,3.5));
 +
label("B",(0,3.5),W);
 +
label("$x$",(0,4.25),W);
 +
label("$h$",(0,1),W);
 +
label("$d$",(4,0),S);</asy>
 +
 +
<math> \textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h </math>
 +
 
== Solution 1==
 
== Solution 1==
  

Latest revision as of 18:31, 2 May 2022

Problem 35

In the right triangle shown the sum of the distances $BM$ and $MA$ is equal to the sum of the distances $BC$ and $CA$. If $MB = x, CB = h$, and $CA = d$, then $x$ equals:

[asy] defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; draw((0,0)--(8,0)--(0,5)--cycle); label("C",(0,0),SW); label("A",(8,0),SE); label("M",(0,5),N); dot((0,3.5)); label("B",(0,3.5),W); label("$x$",(0,4.25),W); label("$h$",(0,1),W); label("$d$",(4,0),S);[/asy]

$\textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h$

Solution 1

The question states that \[h+d = x+\sqrt{(x+h)^2+d^2}\]

We move $x$ to the left: \[h+d-x = \sqrt{(x+h)^2+d^2}\]

We square both sides: \[h^2 + d^2 + x^2 - 2xh - 2xd + 2hd = x^2 + 2xh + h^2 + d^2\]

Cancelling and moving terms, we get: \[4xh + 2xd = 2hd\]

Factoring $x$: \[x(4h+2d) = 2hd\]

Isolating for $x$: \[x=\frac{2hd}{4h+2d}=\frac{hd}{2h+d}\]

Therefore, the answer is $\fbox{A}$


Solution 2

Realize that a 3 - 4 - 5 triangle satisfies these requirements. Checking the answer choices, $\fbox{A}$ is the correct solution.

See Also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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=1954_AHSME_Problems/Problem_35&oldid=173741"