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Difference between revisions of "1993 AHSME Problems/Problem 27"
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== Solution ==
 
== Solution ==
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<math>\fbox{B}</math>
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 +
<asy>
 +
draw(circle((4,1),1),black+linewidth(.75));
 +
draw((0,0)--(8,0)--(8,6)--cycle,black+linewidth(.75));
 +
draw((3,1)--(7,1)--(7,4)--cycle,black+linewidth(.75));
 +
draw((3,1)--(3,0),black+linewidth(.75));
 +
draw((3,1)--(2.4,1.8),black+linewidth(.75));
 +
draw((7,1)--(8,1),black+linewidth(.75));
 +
draw((7,1)--(7,0),black+linewidth(.75));
 +
draw((7,4)--(6.4,4.8),black+linewidth(.75));
 +
MP("A",(0,0),SW);MP("B",(8,0),SE);MP("C",(8,6),NE);MP("P",(4,1),NE);MP("E",(7,1),NE);MP("D",(3,1),SW);MP("G",(3,0),SW);MP("H",(2.4,1.8),NW);MP("F",(7,4),NE);MP("I",(6.4,4.8),NW);
 +
MP("8",(4,0),S);MP("6",(8,3),E);MP("10",(4,3),NW);
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dot((4,1));dot((7,1));dot((3,1));dot((7,4));
 +
</asy>
 +
 
 +
Start by considering the triangle traced by <math>P</math> as the circle moves around the triangle. It turns out this triangle is similar to the <math>6-8-10</math> triangle (Proof: Realize that the slope of the line made while the circle is on <math>AC</math> is the same as line <math>AC</math> and that it makes a right angle when the circle switches from being on <math>AB</math> to <math>BC</math>). Then, drop the perpendiculars as shown.
 +
 
 +
Since the smaller triangle is also a <math>6-8-10 = 3-4-5</math> triangle, we can label the sides <math>EF, </math>  <math>CE, </math> and <math>DF</math> as <math>3x, 4x,</math> and <math>5x</math> respectively. Now, it is clear that <math>GB = DE + 1 = 4x + 1</math>, so <math>AH = AG = 8 - GB = 7 - 4x</math> since <math>AH</math> and <math>AG</math> are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get <math>CI = 5 - 3x</math>. Finally, since we have <math>HI = DF = 5x</math>, we have <math>AC = 10 = (7 - 4x) + (5x) + (5 - 3x) = 12 - 2x</math>, so <math>x = 1</math> and <math>3x + 4x + 5x = \fbox{(B) 12}</math>
 +
 
 +
-Solution by Someonenumber011
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1993|num-b=26|num-a=28}}   
 
{{AHSME box|year=1993|num-b=26|num-a=28}}   
  
−
[[Category:Introductory Algebra Problems]]
+
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:10, 18 July 2020

Problem

[asy] draw(circle((4,1),1),black+linewidth(.75)); draw((0,0)--(8,0)--(8,6)--cycle,black+linewidth(.75)); MP("A",(0,0),SW);MP("B",(8,0),SE);MP("C",(8,6),NE);MP("P",(4,1),NW); MP("8",(4,0),S);MP("6",(8,3),E);MP("10",(4,3),NW); MP("->",(5,1),E); dot((4,1)); [/asy] The sides of $\triangle ABC$ have lengths $6,8,$ and $10$. A circle with center $P$ and radius $1$ rolls around the inside of $\triangle ABC$, always remaining tangent to at least one side of the triangle. When $P$ first returns to its original position, through what distance has $P$ traveled?

$\text{(A) } 10\quad \text{(B) } 12\quad \text{(C) } 14\quad \text{(D) } 15\quad \text{(E) } 17$

Solution

[asy] draw(circle((4,1),1),black+linewidth(.75)); draw((0,0)--(8,0)--(8,6)--cycle,black+linewidth(.75)); draw((3,1)--(7,1)--(7,4)--cycle,black+linewidth(.75)); draw((3,1)--(3,0),black+linewidth(.75)); draw((3,1)--(2.4,1.8),black+linewidth(.75)); draw((7,1)--(8,1),black+linewidth(.75)); draw((7,1)--(7,0),black+linewidth(.75)); draw((7,4)--(6.4,4.8),black+linewidth(.75)); MP("A",(0,0),SW);MP("B",(8,0),SE);MP("C",(8,6),NE);MP("P",(4,1),NE);MP("E",(7,1),NE);MP("D",(3,1),SW);MP("G",(3,0),SW);MP("H",(2.4,1.8),NW);MP("F",(7,4),NE);MP("I",(6.4,4.8),NW); MP("8",(4,0),S);MP("6",(8,3),E);MP("10",(4,3),NW); dot((4,1));dot((7,1));dot((3,1));dot((7,4)); [/asy]

Start by considering the triangle traced by $P$ as the circle moves around the triangle. It turns out this triangle is similar to the $6-8-10$ triangle (Proof: Realize that the slope of the line made while the circle is on $AC$ is the same as line $AC$ and that it makes a right angle when the circle switches from being on $AB$ to $BC$). Then, drop the perpendiculars as shown.

Since the smaller triangle is also a $6-8-10 = 3-4-5$ triangle, we can label the sides $EF,$ $CE,$ and $DF$ as $3x, 4x,$ and $5x$ respectively. Now, it is clear that $GB = DE + 1 = 4x + 1$, so $AH = AG = 8 - GB = 7 - 4x$ since $AH$ and $AG$ are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get $CI = 5 - 3x$. Finally, since we have $HI = DF = 5x$, we have $AC = 10 = (7 - 4x) + (5x) + (5 - 3x) = 12 - 2x$, so $x = 1$ and $3x + 4x + 5x = \fbox{(B) 12}$

-Solution by Someonenumber011

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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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