AMC 10 · 2004 · #22

Grade 8 geometry-2d

tangent-circlespythagorean-theoremcoordinate-geometry convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

Square ABCD has side length 2. A semicircle sits inside the square with its flat diameter along side AB. A line is drawn from corner C so that it just grazes the semicircle (touches it at one point) and then meets side AD at a point E. Find the length of segment CE.

Pick an answer.

(A)

$\frac{2+\sqrt{5}}{2}$

(B)

$\sqrt{5}$

(C)

$\sqrt{6}$

(D)

$\frac{5}{2}$

(E)

$5-\sqrt{5}$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Put the square on a grid so the semicircle's center and radius are exact numbers. The center sits one unit from each vertical side, which is exactly the radius, so both side BC and side AD already touch the circle. That makes CE the third tangent line in the picture, and the two-tangent rule (two tangent segments drawn from the same outside point are equal) lets the length CE be split into two pieces that copy known lengths. Naming the one free length AE then turns the right triangle on the left of the square into a single equation.

1STEP 1

Place the square on a grid

Put A at (0,0), B at (2,0), C at (2,2), D at (0,2); the semicircle's center M is (1,0) with radius 1.

A=(0,0), B=(2,0), C=(2,2), D=(0,2), M=(1,0), r=1

2STEP 2

Spot that two sides already touch the circle

M sits exactly 1 unit (one radius) from line x = 2 and from line x = 0, so sides BC and AD are both tangent, at B and at A.

dist(M, x=2) = 1 = r, dist(M, x=0) = 1 = r

3STEP 3

Use the two-tangent rule from C and from E

Two tangents from one outside point are equal. With F the touch point on CE, CF = CB = 2 and EF = EA.

CF = CB = 2, EF = EA

4STEP 4

Write CE as 2 plus AE

F lies between C and E, so CE = CF + FE. Writing e for the unknown AE gives CE = 2 + e.

CE = CF + FE = 2 + EA = 2 + e

5STEP 5

Get a second expression for CE with the Pythagorean theorem

In right triangle CDE the legs are CD = 2 and DE = 2 - e, so CE² = 2² + (2 - e)².

CE² = CD² + DE² = 2² + (2-e)²

6STEP 6

Solve for AE, then read off CE

Squaring gives (2+e)² = 4 + (2-e)², so e = 1/2 and CE = 2 + 1/2 = 5/2, choice (D).

(2+e)² = 4 + (2-e)² → 4+4e = 8-4e → e=1/2, CE = 2+1/2 = 5/2

Answer

5/2

Check the length directly from the corners: E = (0, 1/2) and C = (2,2), so CE squared is 2² + (3/2)² = 4 + 9/4 = 25/4, and CE = 5/2. That matches. It is also sensible that CE is a bit longer than the side of the square (2) but shorter than the diagonal (about 2.83), and 2.5 sits right in that range. As a sanity check on the tangency, the tangent-length pieces add up: CB = 2 and EA = 1/2 give 2.5 as well.

💡Key takeaway

When a line just grazes a circle, split it at the touch point and copy the equal tangent lengths, then let the Pythagorean theorem close the loop.

Place the square on a grid
Spot that two sides already touch the circle
Use the two-tangent rule from C and from E
Write CE as 2 plus AE
Get a second expression for CE with the Pythagorean theorem
Solve for AE, then read off CE