AMC 10 · 2013 · #23

Grade 7 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

power-of-a-pointisosceles-trianglepolygon-inequality convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: power-of-a-point

📏 Long solution 💡 3 insights

Problem

In $\triangle ABC$ , $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2013 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: In triangle ABC the side AB is 86 and the side AC is 97. A circle is centered at A with radius 86, the same length as AB. This circle crosses side BC at B and at one more point X. The two pieces of the side, BX and CX, both turn out to be whole numbers. Find the length of the whole side BC.

Givens: AB = 86 and AC = 97.; The circle has center A and radius AB = 86.; The circle meets segment BC at B and at a second point X, so X lies on BC.; The lengths BX and CX are both integers.

Unknowns: The length of BC, which equals BX + CX.

Understand

Givens: AB = 86 and AC = 97.; The circle has center A and radius AB = 86.; The circle meets segment BC at B and at a second point X, so X lies on BC.; The lengths BX and CX are both integers.

Plan

Primary tool: #4 Introduce a Variable

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

The two side pieces are hidden, so name them with letters. A circle through B and X lets me turn the side lengths into one clean product using the Power of a Point idea. That product is a fixed number, so I factor it to list the possible values of BC, then use the triangle inequality to erase every value but one.

Execute — Answer: D

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

Draw the circle and mark radii

Sketch triangle ABC and the circle centered at A with radius 86.
Every point on the circle sits 86 away from A, so AB = 86 and AX = 86.
Since AC = 97 is longer than the radius, C is outside the circle.
The straight line from C through the center A pokes the circle at a near point 97 - 86 = 11 away from C and a far point 97 + 86 = 183 away from C.

$$AX = AB = 86,\quad C\text{ outside since }AC = 97 > 86,\quad \text{line }CA\text{ meets circle at distances }97-86=11\text{ and }97+86=183\text{ from }C$$

💡 Everything on a circle is exactly one radius from the center, so B and X are both 86 from A.

#4 Introduce a Variable 6.EE.B.6 Step 2

Name the two side pieces

Let CX = x and BX = y, where x and y are positive whole numbers.
Then the full side is BC = x + y.
Notice that CX = x is the shorter piece from C to X, and CB = x + y is the longer distance from C all the way to B, so CB is bigger than CX.

$$CX = x,\quad BX = y,\quad BC = x + y,\quad CB > CX$$

💡 Giving the unknown pieces names lets the geometry clue become an equation.

#4 Introduce a Variable 7.EE.B.4 Step 3

Use Power of a Point at C

Draw two straight lines out of C that both cut the circle.
One line is CB: it enters at X and leaves at B, at distances x and x + y from C.
The other line is CA: it cuts the circle at distances 11 and 183 from C.
The Power of a Point rule says that for any point, the product of the two crossing distances is the same on every line through it.
So x times (x + y) equals 11 times 183.

$$CX \cdot CB = x(x+y) = 11 \cdot 183 = 2013$$

💡 From one outside point, every secant line hits the circle with the same product of distances.

#4 Introduce a Variable 4.OA.B.4 Step 4

Factor 2013 to list the options

Break 2013 into prime factors: 2013 = 3 * 11 * 61.
I need two factors whose product is 2013, where the smaller factor is x = CX and the larger factor is x + y = BC (larger because y is positive).
The factor pairs are (1, 2013), (3, 671), (11, 183), and (33, 61).
Each pair gives a possible BC of 2013, 671, 183, or 61.

$$2013 = 3 \cdot 11 \cdot 61,\quad (x,\; x+y) \in \{(1,2013),\,(3,671),\,(11,183),\,(33,61)\}$$

💡 The product is locked at 2013, so BC has to be one of its factors.

#3 Eliminate Possibilities 7.G.A.2 Step 5

Erase impossible lengths

The three points A, B, C form a real triangle, so BC must fit the triangle inequality: it must be less than AB + AC = 183 and more than AC - AB = 11.
Check the choices: 2013 and 671 are far too big, and 183 would flatten the triangle into a straight line, so it is not allowed.
Only 61 sits strictly between 11 and 183.
That gives CX = 33 and BX = 61 - 33 = 28, both whole numbers as required.
So BC = 61, and the answer is (D).

$$11 < BC < 183 \;\Rightarrow\; BC = 61,\quad CX = 33,\quad BX = 28$$

💡 A side of a triangle can never reach or pass the sum of the other two sides.

[1] #1 7.G.B.4 Sketch triangle ABC and the circle centered at A with radius 86. Every point on

[2] #4 6.EE.B.6 Let CX = x and BX = y, where x and y are positive whole numbers. Then the full s

[3] #4 7.EE.B.4 Draw two straight lines out of C that both cut the circle. One line is CB: it en

[4] #4 4.OA.B.4 Break 2013 into prime factors: 2013 = 3 * 11 * 61. I need two factors whose prod

[5] #3 7.G.A.2 The three points A, B, C form a real triangle, so BC must fit the triangle inequ

Review

Reasonableness: Test BC = 61. The pieces are CX = 33 and BX = 28, which add to 61 and are both whole numbers. The product 33 * 61 = 2013 matches 11 * 183, so the Power of a Point equation holds. The three sides 86, 97, 61 obey the triangle inequality, since 61 < 86 + 97, 86 < 61 + 97, and 97 < 61 + 86. Everything checks, so (D) is solid.

Alternative: Drop a perpendicular from A to line BC, meeting it at M. Because AB = AX = 86, triangle ABX is isosceles, so M is the midpoint of chord BX and BM = MX. Writing the Pythagorean theorem for the right triangles ABM and ACM and subtracting the two equations relates CX and BX to 97^2 - 86^2 = 2013; the same integer search then forces BX = 28 and CX = 33, giving BC = 61.

CCSS standards used (min grade 7)

7.G.B.4 Know the formulas for area and circumference of a circle (Understanding that every point on the circle, including B and X, is one radius (86) from the center A)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the unknown side pieces CX = x and BX = y and writing BC = x + y)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Turning the Power of a Point relation into the equation x(x + y) = 2013)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Factoring 2013 = 3 * 11 * 61 and listing the factor pairs that BC could be)
7.G.A.2 Draw geometric shapes with given conditions including triangles (Applying the triangle inequality 11 < BC < 183 to eliminate every option except 61)

⭐ From one point outside a circle, every straight cut gives the same product of distances, so name the pieces, get one product, and factor it.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Draw the circle and mark radii

Name the two side pieces

Use Power of a Point at C

Factor 2013 to list the options

Erase impossible lengths

Review

CCSS standards used (min grade 7)

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