🇺🇸 AMC 8 ⇄ switch contest
1996 AMC 8 Stretch

Problem 3

Problem 3 · AMC 8 Stretch Stretch
Geometry & Measurement visual-representationaccount-for-all-possibilitiespythagorean-theorem
Three circles all touch a straight line, and each circle touches the other two. Two of the circles are the same size, and the third (smaller) circle has a radius of 3. All three circles sit on the same side of the line. What is the radius of the two matching circles?
Three mutually tangent circles on a common tangent line L R S 12 9 15 common tangent line
Show answer
Answer: r = 12
Show hints
Hint 1 of 4
Everything depends on a clear picture. Draw the line flat (horizontal). A circle that touches the line sits with its center straight above the touching point, at a height equal to its radius.
Still stuck? Show hint 2 →
Hint 2 of 4
By symmetry, the little circle (radius 3) sits down low between the two big circles (call their radius r). Connect the center of a big circle to the center of the small circle. When two circles touch on the outside, the distance between their centers equals the SUM of the radii: here that is r + 3.
Still stuck? Show hint 3 →
Hint 3 of 4
Make a right triangle. From the big circle's center, the up-and-down leg is the difference in heights, r - 3. The slanted line connecting the two centers (the hypotenuse) is r + 3. You just need the flat (horizontal) leg.
Show solution
Approach: Place centers at height = radius, then use the Pythagorean theorem on the center triangle
  1. Lay the line flat; each circle's center is directly above its touch point, at a height equal to its radius. Put the small circle (radius 3) in the middle at height 3, and the two matching circles (radius r) on either side at height r.
  2. The two big circles touch each other with the small circle exactly between them, so the horizontal distance from a big center to the small center is r.
  3. A big circle touches the small circle externally, so the distance between those centers is \(r+3\); that is the hypotenuse of a right triangle with horizontal leg \(r\) and vertical leg \(r-3\).
  4. Pythagoras: \(r^2+(r-3)^2=(r+3)^2\Rightarrow r^2+r^2-6r+9=r^2+6r+9\Rightarrow r^2-6r=6r\Rightarrow r^2=12r\Rightarrow r=12\).
  5. Check: legs 12 and 9 give hypotenuse \(\sqrt{12^2+9^2}=\sqrt{225}=15=12+3\). So \(r=12\). (The problem says all circles are on the same side of the line; that assumption is what makes the answer unique.)
Mark: · log in to save
↑