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

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==Solution==
 
==Solution==
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[asy]
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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>.  
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pair A, B, C, E, P;
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A=(-20, 0);
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B=origin;
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C=(0,21);
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E=(-21, 20);
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P=extension(B,E, A, C);
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draw(A--B--C--cycle);
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draw(B--P);
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dot("<math>A</math>", A, SW);
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dot("<math>B</math>", B, SE);
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dot("<math>C</math>", C, NE);
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dot("<math>P</math>", P, S);
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[/asy]
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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.
 
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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(asymptote diagram added by elements2015)
  
 
==See Also==
 
==See Also==

Revision as of 18:47, 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] 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)

See Also

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

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