AMC 10 · 2021 · #22

Grade 10 geometry-2d
coordinate-geometrycircle-equationpythagorean-theoremtangent-circles convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometrypythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
A triangle has sides 6, 8, and 10. One circle touches the line of one side at a vertex without crossing it and also runs through another vertex. A second circle touches the line of another side at a different vertex and runs through the remaining one. Find the distance between the two centres.

Pick an answer.

(A)
$\frac{23}{8}$
(B)
$\frac{29}{10}$
(C)
$\frac{35}{12}$
(D)
$\frac{73}{25}$
(E)
3
How to solve
Strategy Convert to Algebra

Two circles are never given by their centers here, so chasing angles is slow. But the triangle is right-angled at C, and the two tangent lines are exactly the two legs CB and CA. Put C at the origin with the legs along the axes and the whole picture becomes coordinates. Then each tangency does heavy lifting: a radius must meet its tangent line at a right angle, so O is forced onto the vertical line through B and P onto the horizontal line through A. Each center therefore has only ONE unknown coordinate left. The 'passes through' condition supplies one equation for that one unknown, and the squared term cancels, leaving plain linear algebra. Finding O and finding P are two separate small problems with the same shape, and once both centers are known the distance formula finishes it.

1STEP 1

Spot the right angle, set coordinates

The sides form a right triangle.

6² + 8² = 100 = 10² → ∠ C = 90°; C=(0,0), B=(6,0), A=(0,8)
2STEP 2

Tangency pins O to one line

Tangency pins the centre to one line.

OB ⊥ BC → O = (6, k), r_O = OB = k
3STEP 3

Passing through A gives k

Passing through a vertex fixes the position.

36 + (k-8)² = k² ⟹ 100 = 16k ⟹ k = 25/4, O = (6, 25/4)
4STEP 4

Repeat the same two moves for P

The second circle repeats the same two moves.

P = (h, 8), (h-6)² + 8² = h² ⟹ 12h = 100 ⟹ h = 25/3, P = (25/3, 8)
5STEP 5

Distance between the centers

Measuring gives thirty-five twelfths.

OP = √((7/3)² + (7/4)²) = √(1225/144) = 35/12
Answer
35/12
Check the centers first. For O = (6, 25/4): OB = 25/4 = 6.25 and OA = √(36 + (8 - 25/4)²) = √(36 + (7/4)²) = √(625/16) = 25/4, equal as required, and O sits directly above B so the circle really does touch the x-axis there. For P = (25/3, 8): PA = 25/3 and PB = √((7/3)² + 64) = √(625/9) = 25/3, equal again, and P sits level with A so that circle touches the y-axis at A. Now the size of OP: 35/12 ≈ 2.92. Both centers are near the corner region of a triangle with legs 6 and 8, and the gaps 7/3 ≈ 2.33 and 7/4 = 1.75 are each smaller than a leg, so a separation just under 3 is sensible. The five choices are all clustered between 23/8 = 2.875 and 3, so a rough sketch cannot separate them, but the exact value 35/12 matches (C) alone.
💡Key takeaway

When a circle touches a line at a named point, its center must sit straight out from that point, so tangency alone knocks a whole unknown out of the problem.

  • Spot the right angle, set coordinates
  • Tangency pins O to one line
  • Passing through A gives k
  • Repeat the same two moves for P
  • Distance between the centers