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 with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E$ . What is the length of $\overline{CE}$ ?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: ABCD is a square with side length 2.; The semicircle has diameter AB, so its center is the midpoint of AB and its radius is 1.; The line CE is tangent to the semicircle, touching it at exactly one point.; E lies on side AD.

Unknowns: The length of segment CE.

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #4 Introduce a Variable, #7 Identify Subproblems

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.

Execute — Answer: D

#1 Draw a Diagram 6.G.A.3 Step 1

Place the square on a grid

Put A at the origin, B at (2,0), C at (2,2), and D at (0,2).
The semicircle has diameter AB, so its center M is the midpoint of AB at (1,0) and its radius is 1.
The center is 1 unit to the right of side AD (the line x = 0) and 1 unit to the left of side BC (the line x = 2), and 1 is exactly the radius.

$$A=(0,0),\; B=(2,0),\; C=(2,2),\; D=(0,2),\quad M=(1,0),\; r=1$$

💡 Fixing the corners as coordinates turns every distance in the picture into arithmetic.

#1 Draw a Diagram 7.G.B.4 Step 2

Spot that two sides already touch the circle

Side BC runs straight up the line x = 2, and its distance from the center M=(1,0) is exactly 1, the radius.
So BC touches the semicircle at one point, meaning BC is tangent, and that touch point is B itself.
The same reasoning makes side AD (the line x = 0) tangent to the circle, touching at A.

$$\text{dist}(M,\, x=2) = 1 = r,\qquad \text{dist}(M,\, x=0) = 1 = r$$

💡 A line one radius away from the center brushes the circle at a single point, so it is a tangent.

#7 Identify Subproblems 8.G.A.2 Step 3

Use the two-tangent rule from C and from E

Two tangent segments drawn from the same outside point to a circle have equal length, because they form two matching right triangles sharing the line to the center.
From corner C, two tangents leave: side CB and the line CE.
Let F be the point where CE grazes the circle.
Then CF equals CB, which is 2.
From E, two tangents leave: the part EA of side AD and the segment EF.
So EF equals EA.

$$CF = CB = 2,\qquad EF = EA$$

💡 Two tangent lines from one point reach the circle equally far, like two equal strings pulled tight from the same hand.

#4 Introduce a Variable 6.EE.B.6 Step 4

Write CE as 2 plus AE

The tangent point F sits on segment CE between C and E, so CE is the sum of its two parts CF and FE.
Replacing CF with 2 and FE with EA gives CE equal to 2 plus AE.
Name the still-unknown length AE with the letter e, so CE = 2 + e.

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

💡 Break the mystery segment at the touch point and each piece copies a length you already control.

#7 Identify Subproblems 8.G.B.7 Step 5

Get a second expression for CE with the Pythagorean theorem

Look at the right triangle CDE, with the right angle at corner D.
Its top leg CD is the full side of the square, length 2.
Its left leg DE is what is left of side AD above E: since AE = e and AD = 2, the piece DE is 2 minus e.
The hypotenuse is CE, so by the Pythagorean theorem CE squared equals 2 squared plus (2 minus e) squared.

$$CE^2 = CD^2 + DE^2 = 2^2 + (2-e)^2$$

💡 The same segment CE now has two descriptions, and forcing them to agree pins down e.

#4 Introduce a Variable 8.EE.C.7 Step 6

Solve for AE, then read off CE

Set the two expressions for CE equal by squaring the first one: (2 + e) squared equals 4 plus (2 minus e) squared.
Expanding both sides, the e-squared terms cancel and the constant 4 cancels, leaving 4e on the left equal to 4 minus 4e plus 4 on the right, that is 8 minus 4e.
So 8e = 4, giving e = 1/2.
Then CE = 2 + 1/2 = 5/2, which is choice (D).

$$(2+e)^2 = 4 + (2-e)^2 \;\Rightarrow\; 4+4e = 8-4e \;\Rightarrow\; e=\tfrac12,\quad CE = 2+\tfrac12 = \tfrac52$$

💡 One equation with one unknown drops straight out once both faces of CE are written down.

[1] #1 6.G.A.3 Put A at the origin, B at (2,0), C at (2,2), and D at (0,2). The semicircle has

[2] #1 7.G.B.4 Side BC runs straight up the line x = 2, and its distance from the center M=(1,0

[3] #7 8.G.A.2 Two tangent segments drawn from the same outside point to a circle have equal le

[4] #4 6.EE.B.6 The tangent point F sits on segment CE between C and E, so CE is the sum of its

[5] #7 8.G.B.7 Look at the right triangle CDE, with the right angle at corner D. Its top leg CD

[6] #4 8.EE.C.7 Set the two expressions for CE equal by squaring the first one: (2 + e) squared

Review

Reasonableness: Check the length directly from the corners: E = (0, 1/2) and C = (2,2), so CE squared is 2^2 + (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.

Alternative: Skip coordinates and use only tangent lengths. Since BC and AD are tangent (each side is one radius from the center), CE = CB + AE = 2 + AE by the two-tangent rule. Then in right triangle CDE, CD = 2 and DE = 2 - AE, so (2 + AE)^2 = 2^2 + (2 - AE)^2; solving gives AE = 1/2 and CE = 5/2.

CCSS standards used (min grade 8)

6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Placing the square's vertices on a grid so the semicircle's center (1,0) and radius 1 are exact.)
7.G.B.4 Know the formulas for area and circumference of a circle (Using the radius to see that sides BC and AD, each one radius from the center, are tangent to the semicircle.)
8.G.A.2 Understand that a two-dimensional figure is congruent to another using transformations (Justifying the two-tangent rule (CF = CB and EF = EA) through the matching right triangles from an external point.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming AE as e and writing CE = 2 + e.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Writing CE squared as 2 squared plus (2 - e) squared in right triangle CDE.)
8.EE.C.7 Solve linear equations in one variable (Setting the two expressions for CE equal and solving 8e = 4 to get e = 1/2.)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

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

Review

CCSS standards used (min grade 8)

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