Difference between revisions of "2017 UNCO Math Contest II Problems/Problem 1"

(Solution)
m
 
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<asy>
 
<asy>
 
pair A=24*dir(40),B=15*dir(220);
 
pair A=24*dir(40),B=15*dir(220);
 +
pair C1=(0,0),C2=(52,0);
  
pair C1=(0,0),C2=(52,0);
 
 
draw(circle(C1,24),black);
 
draw(circle(C1,24),black);
 
draw(circle(C2,15),black);
 
draw(circle(C2,15),black);
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== Solution ==
 
== Solution ==
<math>12</math>
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===Diagram===
 +
<asy>
 +
pair A=24*dir(40), B=15*dir(220);
 +
pair C1=(0,0),C2=(52,0);
 +
pair [] x=intersectionpoints(C1--C2, A--(B+C2));
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pair P=x[0];
 +
 
 +
draw(circle(C1,24),black);
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draw(circle(C2,15),black);
 +
 
 +
draw(C1--C2,dot);
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draw(A--(B+C2), dot);
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draw(P--P, dot);
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draw(C1--A);
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draw(C2--(B+C2));
 +
 
 +
label("A",A,NE);
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label("B",B+C2,SW);
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label("O1",C1,S);
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label("O2",C2,N);
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label("P",P,SW);
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</asy>
 +
===Solution 1===
 +
By [[similar triangles]], <math>O_1P</math> is <math>\frac{24}{39}\cdot 52</math> and <math>O_2P</math> is <math>\frac{15}{39}*52</math>. Their difference is <math>\frac{9}{39}\cdot 52</math>, or <math>\boxed{12}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 18:16, 16 January 2023

Problem

[asy] pair A=24*dir(40),B=15*dir(220); pair C1=(0,0),C2=(52,0);  draw(circle(C1,24),black); draw(circle(C2,15),black);  draw(C1--C2,dot); draw(A--(B+C2)); [/asy]

A circle has radius 24, a second circle has radius 15, and the centers of the two circles are 52 units apart. A line tangent to both circles crosses the line connecting the two centers at a point P between the two centers. How much farther is P from the center of the bigger circle than it is from the center of the smaller circle?

Solution

Diagram

[asy] pair A=24*dir(40), B=15*dir(220); pair C1=(0,0),C2=(52,0); pair [] x=intersectionpoints(C1--C2, A--(B+C2)); pair P=x[0];  draw(circle(C1,24),black); draw(circle(C2,15),black);  draw(C1--C2,dot); draw(A--(B+C2), dot); draw(P--P, dot); draw(C1--A); draw(C2--(B+C2));  label("A",A,NE); label("B",B+C2,SW); label("O1",C1,S); label("O2",C2,N); label("P",P,SW); [/asy]

Solution 1

By similar triangles, $O_1P$ is $\frac{24}{39}\cdot 52$ and $O_2P$ is $\frac{15}{39}*52$. Their difference is $\frac{9}{39}\cdot 52$, or $\boxed{12}$

See also

2017 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions