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Difference between revisions of "1978 AHSME Problems/Problem 26"

(Created page with "==Problem== [asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)...")
 
(→‎Solution)
 
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==Problem==
 
==Problem==
−
[asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram [/asy]
+
<asy>
 +
size(100);
 +
real a=4, b=3;
 +
//
 +
import cse5;
 +
pathpen=black;
 +
pair A=(a,0), B=(0,b), C=(0,0);
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D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle);
 +
pair X=IP(B--A,(0,0)--(b,a));
 +
D(CP((X+C)/2,C));
 +
D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0))));
 +
//Credit to chezbgone2 for the diagram
 +
</asy>
 +
 
  
 
In <math>\triangle ABC, AB = 10~ AC = 8</math> and <math>BC = 6</math>. Circle <math>P</math> is the circle with smallest radius which passes through <math>C</math> and is tangent to <math>AB</math>. Let <math>Q</math> and <math>R</math> be the points of intersection, distinct from <math>C</math> , of circle <math>P</math> with sides <math>AC</math> and <math>BC</math>, respectively. The length of segment <math>QR</math> is
 
In <math>\triangle ABC, AB = 10~ AC = 8</math> and <math>BC = 6</math>. Circle <math>P</math> is the circle with smallest radius which passes through <math>C</math> and is tangent to <math>AB</math>. Let <math>Q</math> and <math>R</math> be the points of intersection, distinct from <math>C</math> , of circle <math>P</math> with sides <math>AC</math> and <math>BC</math>, respectively. The length of segment <math>QR</math> is
  
 
<math>\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad  \textbf{(E) }3\sqrt{3}</math>
 
<math>\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad  \textbf{(E) }3\sqrt{3}</math>
 +
 
==Solution==
 
==Solution==
−
<math>\fbox{B}</math>
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We know that triangle <math>RCQ</math> is similar to triangle <math>ABC</math>. We draw a line to point <math>D</math> on hypotenuse <math>AB</math> such that <math>\angle QDR</math> is <math>90 ^\circ</math> and that <math>RDQC</math> is a rectangle. Since triangle <math>RCQ</math> is similar to triangle <math>ABC</math>, let <math>RC</math> be <math>4x</math> and <math>RD/CQ</math> be <math>3x</math>. Now we have line segment <math>AQ</math> = <math>8-3x</math>, and line segment <math>RB</math> = <math>6-4x</math>. Since <math>BD + DA = AB</math>, we use simple algebra and Pythagorean Theorem to get <math>\sqrt {(3x)^2 + (6-4x)^2}</math> + <math>\sqrt {(4x)^2 + (8-3x)^2}</math> = <math>10</math>. Expanding and simplifying gives us <math>\sqrt {25x^2-48x+36}</math> + <math>\sqrt {25x^2-48x+64}</math> = <math>10</math>.
 +
 
 +
Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by  <math>\sqrt {25x^2-48x+36}</math>. Now, we can square both sides and simplify to get <math>0 = 72 - 20 \sqrt{25x^2-48x+36}</math>. Dividing both sides by <math>4</math>, we get <math>18 - 5 \sqrt {25x^2-48x+36}</math> = <math>0</math>. We then add <math>5 \sqrt {25x^2-48x+36}</math> to both sides to get <math>18 = 5 \sqrt {25x^2-48x+36}</math>. Since this is very messy, let <math>25x^2 - 48x = y</math>. Squaring both sides, we get <math>324 = 25y + 900, 25y = -576</math>. Solving for <math>y</math>, we have <math>y = -23.04</math>. Plugging in <math>y</math> as <math>25x^2-48x</math>, we have <math>25x^2-48x+23.04 = 0</math>. Using the quadratic equation, we get <math>\frac {48+0}{50}</math>. Therefore, <math>x = \frac {48}{50}</math>.
 +
 
 +
Remember that our desired answer is the hypotenuse of the triangle <math>3x - 4x - 5x</math>. Since <math>5x</math> is the hypotenuse, our answer is <math>\boxed {(B)4.8}</math>
 +
 
 +
~Arcticturn
 +
 
 +
==See Also==
 +
{{AHSME box|year=1978|num-b=25|num-a=27}} 
 +
{{MAA Notice}}

Latest revision as of 18:28, 6 November 2021

Problem

[asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram [/asy]


In $\triangle ABC, AB = 10~ AC = 8$ and $BC = 6$. Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$. Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$, respectively. The length of segment $QR$ is

$\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }3\sqrt{3}$

Solution

We know that triangle $RCQ$ is similar to triangle $ABC$. We draw a line to point $D$ on hypotenuse $AB$ such that $\angle QDR$ is $90 ^\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$, let $RC$ be $4x$ and $RD/CQ$ be $3x$. Now we have line segment $AQ$ = $8-3x$, and line segment $RB$ = $6-4x$. Since $BD + DA = AB$, we use simple algebra and Pythagorean Theorem to get $\sqrt {(3x)^2 + (6-4x)^2}$ + $\sqrt {(4x)^2 + (8-3x)^2}$ = $10$. Expanding and simplifying gives us $\sqrt {25x^2-48x+36}$ + $\sqrt {25x^2-48x+64}$ = $10$.

Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\sqrt {25x^2-48x+36}$. Now, we can square both sides and simplify to get $0 = 72 - 20 \sqrt{25x^2-48x+36}$. Dividing both sides by $4$, we get $18 - 5 \sqrt {25x^2-48x+36}$ = $0$. We then add $5 \sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \sqrt {25x^2-48x+36}$. Since this is very messy, let $25x^2 - 48x = y$. Squaring both sides, we get $324 = 25y + 900, 25y = -576$. Solving for $y$, we have $y = -23.04$. Plugging in $y$ as $25x^2-48x$, we have $25x^2-48x+23.04 = 0$. Using the quadratic equation, we get $\frac {48+0}{50}$. Therefore, $x = \frac {48}{50}$.

Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$. Since $5x$ is the hypotenuse, our answer is $\boxed {(B)4.8}$

~Arcticturn

See Also

1978 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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