1979 AHSME Problems/Problem 28

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




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 (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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