AMC 10 · 2004 · #18

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, and a semicircle sits inside it with its flat diameter along side AB. A line from corner C just grazes the semicircle 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)

$5-\sqrt{5}$

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

On a grid the semicircle has centre at the midpoint of the base and 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

Two of the square's sides sit exactly a radius away, so both are tangent.

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

3STEP 3

Use the two-tangent rule from C and from E

Equal tangents from each outside point give CF = 2 and EF = EA.

CF = CB = 2, EF = EA

4STEP 4

Write CE as 2 plus AE

So the whole segment is 2 plus AE.

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

5STEP 5

Get a second expression for CE with the Pythagorean theorem

A right triangle at the far corner gives a second expression for the same segment.

CE² = CD² + DE² = 2² + (2-e)²

6STEP 6

Solve for AE, then read off CE

Equating them gives AE as a half, so the segment is 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 5/2 = 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