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1958 AHSME Problems/Problem 50

Problem

In this diagram a scheme is indicated for associating all the points of segment $\overline{AB}$ with those of segment $\overline{A'B'}$, and reciprocally. To described this association scheme analytically, let $x$ be the distance from a point $P$ on $\overline{AB}$ to $D$ and let $y$ be the distance from the associated point $P'$ of $\overline{A'B'}$ to $D'$. Then for any pair of associated points, if $x = a,\, x + y$ equals:

[asy] draw((0,-3)--(0,3),black+linewidth(.75)); draw((1,-2.5)--(5,-2.5),black+linewidth(.75)); draw((3,2.5)--(4,2.5),black+linewidth(.75)); draw((1,-2.5)--(4,2.5),black+linewidth(.75)); draw((5,-2.5)--(3,2.5),black+linewidth(.75)); draw((2.6,-2.5)--(3.6,2.5),black+linewidth(.75)); dot((0,2.5));dot((1,2.5));dot((2,2.5));dot((3,2.5));dot((4,2.5));dot((5,2.5)); dot((0,-2.5));dot((1,-2.5));dot((2,-2.5));dot((3,-2.5));dot((4,-2.5));dot((5,-2.5)); MP("D",(0,2.5),NW);MP("A",(3,2.5),N);MP("P",(3.5,2.5),N);MP("B",(4,2.5),N); MP("D'",(0,-2.5),NW);MP("B'",(1,-2.5),NW);MP("P'",(2.25,-2.5),N);MP("A'",(5,-2.5),NE); MP("0",(0,2.5),SE);MP("1",(1,2.5),SE);MP("2",(2,2.5),SE);MP("3",(3,2.5),SE);MP("4",(4,2.5),SE);MP("5",(5,2.5),SE); MP("0",(0,-2.5),SE);MP("1",(1,-2.5),SE);MP("2",(2,-2.5),SE);MP("3",(3,-2.5),SE);MP("4",(4,-2.5),SE);MP("5",(5,-2.5),SE); [/asy]

$\textbf{(A)}\ 13a\qquad \textbf{(B)}\ 17a - 51\qquad \textbf{(C)}\ 17 - 3a\qquad \textbf{(D)}\ \frac {17 - 3a}{4}\qquad \textbf{(E)}\ 12a - 34$

Solution

$\fbox{}$

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 49 		Followed by
Problem 50
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

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