Difference between revisions of "1954 AHSME Problems/Problem 35"

Revision as of 21:32, 24 April 2020

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 $\fbox{A}$

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

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

@@ Line 19: / Line 19: @@
 Realize that a 3 - 4 - 5 triangle satisfies these requirements. Checking the answer choices, <math>\fbox{A}</math> is the correct solution.
+==See Also==
+{{AHSME 50p box|year=1954|num-b=34|num-a=36}}
+{{MAA Notice}}