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

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==Problem==
 
==Problem==
 +
  
 
Right triangle <math>ABC</math> has leg lengths <math>AB=20</math> and <math>BC=21</math>. Including <math>\overline{AB}</math> and <math>\overline{BC}</math>, how many line segments with integer length can be drawn from vertex <math>B</math> to a point on hypotenuse <math>\overline{AC}</math>?
 
Right triangle <math>ABC</math> has leg lengths <math>AB=20</math> and <math>BC=21</math>. Including <math>\overline{AB}</math> and <math>\overline{BC}</math>, how many line segments with integer length can be drawn from vertex <math>B</math> to a point on hypotenuse <math>\overline{AC}</math>?
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</math>
 
</math>
  
−
==Solution==
+
==Solution 1==
 +
 
 
<asy>
 
<asy>
 
unitsize(4);
 
unitsize(4);
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dot("$B$", B, SE);
 
dot("$B$", B, SE);
 
dot("$C$", C, NE);
 
dot("$C$", C, NE);
−
dot("$P$", P, S);
+
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.
+
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)
  
 
==Solution 2 - Circles==
 
==Solution 2 - Circles==
−
Note that if a circle with an integer radius <math>r</math> centered at vertex <math>B</math> intersects hypotenuse <math>\overline{AB}</math>, the lines drawn from <math>B</math> to the points of intersection are integer lengths. As in the previous solution, the shortest distance <math>14<\overline{BP}<15</math>. As a result, a circle of <math>14</math> will [b]not[/b] reach the hypotenuse and thus does not intersect it. We also know that a circle of radius <math>21</math> intersects the hypotenuse once and a circle of radius <math>\{15, 16, 17, 18, 19, 20 \}</math> intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.
+
Note that if a circle with an integer radius <math>r</math> centered at vertex <math>B</math> intersects hypotenuse <math>\overline{AB}</math>, the lines drawn from <math>B</math> to the points of intersection are integer lengths. As in the previous solution, the shortest distance <math>14<\overline{BP}<15</math>. As a result, a circle of <math>14</math> will <b>not</b> reach the hypotenuse and thus does not intersect it. We also know that a circle of radius <math>21</math> intersects the hypotenuse once and a circle of radius <math>\{15, 16, 17, 18, 19, 20 \}</math> intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.
 +
 
 
<asy>
 
<asy>
 
unitsize(4);
 
unitsize(4);
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dot("$B$", B, SE);
 
dot("$B$", B, SE);
 
dot("$C$", C, NE);
 
dot("$C$", C, NE);
−
dot("$P$", P, S);]
+
dot("$P$", P, S);
−
draw(circle((0,0), 21));
+
draw(arc((0,0),21, 90, 180));
−
draw(circle((0,0), 20));
+
draw(arc((0,0),20, 90, 180));
−
draw(circle((0,0), 19));
+
draw(arc((0,0),19, 90, 180));
−
draw(circle((0,0), 18));
+
draw(arc((0,0),18, 90, 180));
−
draw(circle((0,0), 17));
+
draw(arc((0,0),17, 90, 180));
−
draw(circle((0,0), 16));
+
draw(arc((0,0),16, 90, 180));
−
draw(circle((0,0), 15));
+
draw(arc((0,0),15, 90, 180));
 
</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
 
  
−
==Video Solution 1==
+
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(The Beauty of Math)==
 
https://youtu.be/M22S82Am2zM
 
https://youtu.be/M22S82Am2zM
  
 
~IceMatrix
 
~IceMatrix
  
−
== Video Solution 2 ==
+
== Video Solution 2 by OmegaLearn ==
 
https://youtu.be/4_x1sgcQCp4?t=3790
 
https://youtu.be/4_x1sgcQCp4?t=3790
  

Latest revision as of 18:14, 31 October 2024

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(The Beauty of Math)

https://youtu.be/M22S82Am2zM

~IceMatrix

Video Solution 2 by OmegaLearn

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

~ pi_is_3.14

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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