Difference between revisions of "2018 AMC 10A Problems/Problem 16"

Revision as of 18:50, 3 March 2018

Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$ . Including $\overline{AB}$ and $\overline{BC}$ , how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$ ?

$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }15 \qquad$

Solution

$[asy] unitsize(3); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$P$", P, S); [/asy]$ As the problem has no diagram, we draw a diagram. The hypotenuse has length $29$ . Let $P$ be the foot of the altitude from $B$ to $AC$ . Note that $BP$ is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for $BP=\dfrac{20\cdot 21}{29}$ , which is between $14$ and $15$ .

Let the line segment be $BX$ , with $X$ on $AC$ . As you move $X$ along the hypotenuse from $A$ to $P$ , the length of $BX$ strictly decreases, hitting all the integer values from $20, 19, \dots 15$ (IVT). Similarly, moving $X$ from $P$ to $C$ hits all the integer values from $15, 16, \dots, 21$ . This is a total of $\boxed{(D) 13}$ line segments. (asymptote diagram added by elements2015)

2018 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2018 AMC 10A Problems/Problem 16"

Revision as of 18:50, 3 March 2018

Solution

See Also

@@ Line 11: / Line 11: @@
 ==Solution==
 <asy>
-unitsize(10);
+unitsize(3);
 pair A, B, C, E, P;
 A=(-20, 0);