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1979 AHSME Problems/Problem 17

Problem 17

[asy] size(200); dotfactor=3; pair A=(0,0),B=(1,0),C=(2,0),D=(3,0),X=(1.2,0.7); draw(A--D); dot(A);dot(B);dot(C);dot(D); draw(arc((0.4,0.4),0.4,180,110),arrow = Arrow(TeXHead)); draw(arc((2.6,0.4),0.4,0,70),arrow = Arrow(TeXHead)); draw(B--X,dotted); draw(C--X,dotted); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("x",X,fontsize(5pt)); //Credit to TheMaskedMagician for the diagram [/asy]

Points $A , B, C$, and $D$ are distinct and lie, in the given order, on a straight line. Line segments $AB, AC$, and $AD$ have lengths $x, y$, and $z$, respectively. If line segments $AB$ and $CD$ may be rotated about points $B$ and $C$, respectively, so that points $A$ and $D$ coincide, to form a triangle with positive area, then which of the following three inequalities must be satisfied?

$\textbf{I. }x<\frac{z}{2}\qquad\\ \textbf{II. }y<x+\frac{z}{2}\qquad\\ \textbf{III. }y<\frac{z}{2}\qquad$


$\textbf{(A) }\textbf{I. }\text{only}\qquad \textbf{(B) }\textbf{II. }\text{only}\qquad \textbf{(C) }\textbf{I. }\text{and }\textbf{II. }\text{only}\qquad \textbf{(D) }\textbf{II. }\text{and }\textbf{III. }\text{only}\qquad \textbf{(E) }\textbf{I. },\textbf{II. },\text{and }\textbf{III. }$

Solution

Solution by e_power_pi_times_i

We know that this triangle has lengths of $x$, $y-x$, and $z-y$. Using the Triangle Inequality, we get $3$ inequalities: $2y>z, z>2x, 2x+z>0$. Therefore, we know that $\textbf{I}$ is true and $\textbf{III}$ is false. In $\textbf{II}$, we have to prove $2y<2x+z$. We know that $2y>z$, so we have to prove $z<2y<2x+z$. $2x<z$, so we have to prove that $z<2z$, which is true for all positive $z$. Therefore the answer is $\boxed{\textbf{(C) } \textbf{I. }\text{and }\textbf{II. }\text{only}}$.

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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