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

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dot("$C$", C, NE);
 
dot("$C$", C, NE);
 
dot("$P$", P, S);
 
dot("$P$", P, S);
dot("$R$", (-3,10), N);
+
dot("$R$", (-6,12), N);
 
</asy>
 
</asy>
  

Revision as of 14:02, 16 October 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 1

[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, NW); [/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 $\boxed{(D) 13}$ distinct 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 not 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 two integer lengths (since these circles intersect the hypotenuse twice) 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

Solution 3 - Triangle Inequality

[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); dot("$R$", (-6,12), N); [/asy]

Video Solution 1

https://youtu.be/M22S82Am2zM

~IceMatrix

Video Solution 2

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

~ pi_is_3.14

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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

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