Difference between revisions of "1954 AHSME Problems/Problem 35"
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+ | == 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 17:31, 2 May 2022
Contents
[hide]Problem 35
In the right triangle shown the sum of the distances and is equal to the sum of the distances and . If , and , then equals:
Solution 1
The question states that
We move to the left:
We square both sides:
Cancelling and moving terms, we get:
Factoring :
Isolating for :
Therefore, the answer is
Solution 2
Realize that a 3 - 4 - 5 triangle satisfies these requirements. Checking the answer choices, is the correct solution.
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Problem 36 | |
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All AHSME Problems and Solutions |
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