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

 
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</math>
 
</math>
  
==Solution==
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==Solution 1==
  
 
<asy>
 
<asy>
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dot("$B$", B, SE);
 
dot("$B$", B, SE);
 
dot("$C$", C, NE);
 
dot("$C$", C, NE);
dot("$P$", P, S);
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dot("$P$", P, NW);
 
</asy>
 
</asy>
  
 
As the problem has no diagram, we draw a diagram. The hypotenuse has length <math>29</math>. Let <math>P</math> be the foot of the altitude from <math>B</math> to <math>AC</math>. Note that <math>BP</math> is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for <math>BP=\dfrac{20\cdot  21}{29}</math>, which is between <math>14</math> and <math>15</math>.  
 
As the problem has no diagram, we draw a diagram. The hypotenuse has length <math>29</math>. Let <math>P</math> be the foot of the altitude from <math>B</math> to <math>AC</math>. Note that <math>BP</math> is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for <math>BP=\dfrac{20\cdot  21}{29}</math>, which is between <math>14</math> and <math>15</math>.  
  
Let the line segment be <math>BX</math>, with <math>X</math> on <math>AC</math>. As you move <math>X</math> along the hypotenuse from <math>A</math> to <math>P</math>, the length of <math>BX</math> strictly decreases, hitting all the integer values from <math>20, 19, \dots 15</math> (IVT). Similarly, moving <math>X</math> from <math>P</math> to <math>C</math> hits all the integer values from <math>15, 16, \dots, 21</math>. This is a total of <math>\boxed{(D) 13}</math> line segments.
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Let the line segment be <math>BX</math>, with <math>X</math> on <math>AC</math>. As you move <math>X</math> along the hypotenuse from <math>A</math> to <math>P</math>, the length of <math>BX</math> strictly decreases, hitting all the integer values from <math>20, 19, \dots 15</math> (IVT). Similarly, moving <math>X</math> from <math>P</math> to <math>C</math> hits all the integer values from <math>15, 16, \dots, 21</math>. This is a total of <math>\boxed{(D) 13}</math> distinct line segments.
 
(asymptote diagram added by elements2015)
 
(asymptote diagram added by elements2015)
  
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</asy>
 
</asy>
  
It follows that we can draw circles of radii <math>15, 16, 17, 18, 19,</math> and <math>20,</math> that each contribute <b>two</b> integer lengths from <math>B</math> to <math>\overline{AC}</math> and one circle of radius <math>21</math> that contributes only one such segment. Our answer is then <cmath>6 \cdot 2 + 1 = 13 \implies \boxed{D}</cmath> ~samrocksnature
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It follows that we can draw circles of radii <math>15, 16, 17, 18, 19,</math> and <math>20,</math> that each contribute <b>two</b> integer lengths (since these circles intersect the hypotenuse twice) from <math>B</math> to <math>\overline{AC}</math> and one circle of radius <math>21</math> that contributes only one such segment. Our answer is then <cmath>6 \cdot 2 + 1 = 13 \implies \boxed{D}</cmath> ~samrocksnature
  
 
==Video Solution 1==
 
==Video Solution 1==

Latest revision as of 15:09, 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

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
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All AMC 10 Problems and Solutions

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