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

Revision as of 22:48, 1 February 2021

Problem

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

$[asy] unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot("$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)

Solution 2 - Circles

Note that if a circle with an integer radius $r$ centered at vertex $B$ intersects hypotenuse $\overline{AB}$ , the lines drawn from $B$ to the points of intersection are integer lengths. As in the previous solution, the shortest distance $14<\overline{BP}<15$ . As a result, a circle of $14$ will [b]not[/b] reach the hypotenuse and thus does not intersect it. We also know that a circle of radius $21$ intersects the hypotenuse once and a circle of radius $\{15, 16, 17, 18, 19, 20 \}$ intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this. $[asy] unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$P$", P, S); draw(arc((0,0),21, 90, 180)); draw(arc((0,0),20, 90, 180)); draw(arc((0,0),19, 90, 180)); draw(arc((0,0),18, 90, 180)); draw(arc((0,0),17, 90, 180)); draw(arc((0,0),16, 90, 180)); draw(arc((0,0),15, 90, 180)); [/asy]$ It follows that we can draw circles of radii $15, 16, 17, 18, 19,$ and $20,$ that each contribute [b]two[/b] integer lengths from $B$ to $\overline{AC}$ and one circle of radius $21$ that contributes only one such segment. Our answer is then $6 \cdot 2 + 1 = 13 \implies \boxed{D}$ ~samrocksnature

Video Solution 1

https://youtu.be/M22S82Am2zM

~IceMatrix

Video Solution 2

https://youtu.be/4_x1sgcQCp4?t=3790

~ pi_is_3.14

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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
All AMC 10 Problems and Solutions

Page

Toolbox

Search