Difference between revisions of "2023 AMC 12A Problems/Problem 18"

Revision as of 22:29, 9 November 2023

Problem

Circle $C_1$ and $C_2$ each have radius $1$ , and the distance between their centers is $\frac{1}{2}$ . Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$ . Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$ . What is the radius of $C_4$ ?

[someone pls insert diagram]

$\textbf{(A) } \frac{1}{14} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{10} \qquad \textbf{(D) } \frac{3}{28} \qquad \textbf{(E) } \frac{1}{9}$

Solution 1

With some simple geometry skills, we can find that $C_3$ has a radius of $\frac{3}{4}$ .

Since $C_4$ is internally tangent to $C_1$ , center of $C_4$ , $C_1$ and their tangent point must be on the same line.

Now, if we connect centers of $C_4$ , $C_3$ and $C_1$ / $C_2$ , we get a right angled triangle.

In which we get an equation by pythagorean theorem:

$(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2$

Solving it gives us

$r = \boxed{\textbf{(D) } \frac{3}{28}}$

~lptoggled

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 26: / Line 26: @@
 ==See Also==
-{{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}}
+{{AMC12 box|year=2023|ab=A|num-b=17|num-a=19}}
 {{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "2023 AMC 12A Problems/Problem 18"

Revision as of 22:29, 9 November 2023

Problem

Solution 1

See Also