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Difference between revisions of "1979 AHSME Problems/Problem 28"

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By symmetry across the y-axis the length of the line segment <math>B'C'</math> is <math>1+\sqrt{3(r^2-1)}</math> which is <math>\boxed{D}</math>.
 
By symmetry across the y-axis the length of the line segment <math>B'C'</math> is <math>1+\sqrt{3(r^2-1)}</math> which is <math>\boxed{D}</math>.
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==Solution 2 (Synthetic)==
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Suppose <math>B’B</math> and <math>AC</math> intersect at <math>P</math>. By the Pythagorean Theorem, <math>B’P = \sqrt{r^2 - 1}</math> and by a <math>30-60-90</math> triangle, <math>PB = \sqrt{3}</math>. Using Ptolemy’s Theorem on isosceles trapezoid <math>BCB’C’</math>, we get that <cmath>2(B’C’) + r^2 = (\sqrt{3} + \sqrt{r^2 - 1})^2.</cmath> After a little algebra, we get that <math>B’C’ = \boxed{1 + \sqrt{3(r^2 - 1)}}</math> as desired.
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[[User:Solasky|Solasky]] ([[User talk:Solasky|talk]]) 12:29, 27 May 2023 (EDT)
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1979|num-b=27|num-a=29}}   
 
{{AHSME box|year=1979|num-b=27|num-a=29}}   
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:29, 27 May 2023

Problem 28

[asy] import cse5; pathpen=black; pointpen=black; dotfactor=3; pair A=(1,2),B=(2,0),C=(0,0); D(CR(A,1.5)); D(CR(B,1.5)); D(CR(C,1.5)); D(MP("$A$",A)); D(MP("$B$",B)); D(MP("$C$",C)); pair[] BB,CC; CC=IPs(CR(A,1.5),CR(B,1.5)); BB=IPs(CR(A,1.5),CR(C,1.5)); D(BB[0]--CC[1]); MP("$B'$",BB[0],NW);MP("$C'$",CC[1],NE); //Credit to TheMaskedMagician for the diagram[/asy]

Circles with centers $A ,~ B$, and $C$ each have radius $r$, where $1 < r < 2$. The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$, then length $B'C'$ equals

$\textbf{(A) }3r-2\qquad \textbf{(B) }r^2\qquad \textbf{(C) }r+\sqrt{3(r-1)}\qquad\\ \textbf{(D) }1+\sqrt{3(r^2-1)}\qquad \textbf{(E) }\text{none of these}$


Solution 1 (Coordinate Geometry)

The circles can be described in the cartesian plane as being centered at $(-1,0),(1,0)$ and $(0,\sqrt{3})$ with radius $r$ by the equations

$x^2+(y-\sqrt{3})^2=r^2$

$(x+1)^2+y^2=r^2$

$(x-1)^2+y^2=r^2$.

Solving the first 2 equations gives $x=1-\sqrt{3}\cdot y$ which when substituted back in gives $y=\frac{\sqrt{3}\pm \sqrt{r^2-1}}{2}$.

The larger root $y=\frac{\sqrt{3}+\sqrt{r^2-1}}{2}$ is the point B' described in the question. This root corresponds to $x=-\frac{1+\sqrt{3(r^2-1)}}{2}$.

By symmetry across the y-axis the length of the line segment $B'C'$ is $1+\sqrt{3(r^2-1)}$ which is $\boxed{D}$.

Solution 2 (Synthetic)

Suppose $B’B$ and $AC$ intersect at $P$. By the Pythagorean Theorem, $B’P = \sqrt{r^2 - 1}$ and by a $30-60-90$ triangle, $PB = \sqrt{3}$. Using Ptolemy’s Theorem on isosceles trapezoid $BCB’C’$, we get that \[2(B’C’) + r^2 = (\sqrt{3} + \sqrt{r^2 - 1})^2.\] After a little algebra, we get that $B’C’ = \boxed{1 + \sqrt{3(r^2 - 1)}}$ as desired. Solasky (talk) 12:29, 27 May 2023 (EDT)

See Also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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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