AMC 10 · 2017 · #24

Grade 11 geometry-2d
power-of-a-pointsimilar-trianglescyclic-quadrilaterallaw-of-cosines identify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-trianglespower-of-a-point
📏 Long solution 💡 4 insights
Problem
Parallel lines and a circle build a chain of points from one diagonal. Find a product of two lengths.

Pick an answer.

(A)
17
(B)
$\frac{59 - 5\sqrt{2}}{3}$
(C)
$\frac{91 - 12\sqrt{3}}{4}$
(D)
$\frac{67 - 10\sqrt{2}}{3}$
(E)
18
How to solve
Strategy Identify Subproblems

Neither XF nor XG is easy to find on its own, but the question asks only for their product. Split the work into three independent pieces. The two parallel lines force a chain of similar triangles that expresses XF as a fixed multiple of XC. Power of a Point converts XC · XG into XB · XD, which lives entirely on the diagonal. The only genuine measurement left is BD, and the Law of Cosines pulls that out of the four side lengths. When the pieces are multiplied, XC cancels, so no length along line CX ever has to be computed.

1STEP 1

Draw it and name one length

One letter names the whole diagonal.

BD=q, DX=q/4, BX=3q/4, BY=11q/36
2STEP 2

Measure the gap XY

The given fractions place both marked points.

XY=27q/36-11q/36=4q/9, XY/XD=16/9
3STEP 3

Parallel to AD gives the first similar pair

The first parallel gives a ratio.

△ XAD ∼ △ XEY ⟹ XE/XA=XY/XD=16/9
4STEP 4

Parallel to AC passes the ratio along

The second passes that ratio along.

△ XAC ∼ △ XEF ⟹ XF=16/9 XC
5STEP 5

Power of a Point handles XG

The power of a point handles the circle.

XC · XG=XB · XD=3q/4·q/4=3q²/16
6STEP 6

Multiply so that XC cancels

Multiplying makes one length cancel.

XF · XG=16/9(XC · XG)=16/9·3q²/16=q²/3
7STEP 7

Write BD squared two ways

Two expressions for the diagonal agree.

q²=73-48cos∠ BAD=40+24cos∠ BAD
8STEP 8

Solve for the cosine and finish

That gives 17, choice (A).

cos∠ BAD=11/24, q²=51, XF · XG=51/3=17
Answer
17
Rebuild the figure with actual numbers. Placing B and D a distance √(51)≈7.1414 apart and attaching A with AB=3, AD=8 on one side and C with CB=2, CD=6 on the other, all four vertices land on a single circle of radius about 4.018, which confirms BD²=51 is the value that makes ABCD cyclic. In that figure XF≈7.639 and XG≈2.220, and their product is 17.000. Ptolemy's Theorem is an independent check on the same figure: AC · BD should equal AB · CD+BC · DA=3·6+2·8=34, and it does. The answer choices are deliberately tight, with (B)≈17.31, (C)≈17.55, (D)≈17.62, and (E)=18, so an estimate cannot separate them. The fact that every square root cancels and an integer survives is itself evidence the structure was used rather than approximated.
💡Key takeaway

When a problem asks for a product, look for a way to make the unknown factor cancel instead of chasing it: here every length along line CX disappears and only the diagonal BD is left to find.

  • Draw it and name one length
  • Measure the gap XY
  • Parallel to AD gives the first similar pair
  • Parallel to AC passes the ratio along
  • Power of a Point handles XG
  • Multiply so that XC cancels
  • Write BD squared two ways
  • Solve for the cosine and finish