AMC 10 · 2024 · #19

Grade 11 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
cyclic-quadrilaterallaw-of-cosinesptolemys-theoreminscribed-anglequadratic-equations identify-subproblemsconvert-to-algebra ↑ Prerequisites: law-of-cosinesinscribed-anglequadratic-equations
📏 Medium solution 💡 3 insights
Problem
Four points A, B, C, D lie on one circle, in that order around it, forming quadrilateral ABCD. Three of its four sides are given: BC = 3, CD = 3, and DA = 5. The fourth side AB is not given. The interior angle at D, the one between side DC and side DA, measures 120°. The quadrilateral has two diagonals, AC and BD. Find the length of whichever of those two is shorter.

Pick an answer.

(A)
$\frac{31}7$
(B)
$\frac{33}7$
(C)
5
(D)
$\frac{39}7$
(E)
$\frac{41}7$
How to solve
Strategy Identify Subproblems

A quadrilateral is not a shape with formulas of its own, but a triangle is. So the move is to slice ABCD into triangles along a diagonal and work on triangles only. The slice is not arbitrary: the given 120° angle sits exactly between the two given sides CD and DA, so cutting along AC produces one triangle where two sides and the angle between them are all known. That triangle falls immediately to the Law of Cosines and hands over AC. The other triangle, △ ABC, then has two knowns and one unknown side, and the cyclic condition supplies the missing angle to close it. After that, all four sides and one diagonal are known, and Ptolemy's relation converts that into the second diagonal in one line. The last piece of the plan is the piece most people skip: the problem asks for the shorter diagonal, so both must be computed and compared, never assumed.

1STEP 1

Draw it and choose the cut

Start with triangle ACD, which owns the 120 degrees.

Diagonals: AC, BD. Given: BC = 3, CD = 3, DA = 5, ∠ CDA = 120°. Missing side: AB
2STEP 2

Law of cosines gives AC

It comes out clean: AC is 7.

AC² = CD² + DA² - 2 · CD · DA · cos ∠ CDA = 3² + 5² - 2(3)(5)(-1/2) = 9 + 25 + 15 = 49 ⟹ AC = 7
3STEP 3

The circle fixes the angle at B

Being opposite, B measures 60 degrees.

∠ ABC + ∠ CDA = 1/2(arc ADC) + 1/2(arc ABC) = 1/2(360°) = 180° ⟹ ∠ ABC = 180° - 120° = 60°
4STEP 4

Solve a quadratic for AB

Only the positive root survives: AB is 8.

AC² = AB² + BC² - 2 · AB · BC · cos ∠ ABC ⟹ 49 = x² + 9 - 2(x)(3)(1/2) = x² - 3x + 9 ⟹ x² - 3x - 40 = 0 ⟹ (x - 8)(x + 5) = 0 ⟹ AB = 8
5STEP 5

Ptolemy delivers the second diagonal

Ptolemy gives BD as 39 over 7.

AC · BD = AB · CD + BC · DA ⟹ 7 · BD = 8 · 3 + 3 · 5 = 24 + 15 = 39 ⟹ BD = 39/7
6STEP 6

Compare, do not assume

Measured against 7, the shorter is 39/7.

AC = 7 = 49/7, BD = 39/7, 39 ＜ 49 ⟹ 39/7 ＜ 49/7 ⟹ BD ＜ AC
Answer
39/7
First, a size check the circle itself enforces. The extended Law of Sines in △ ACD gives the diameter as 2R = AC/(sin ∠ ADC) = 7/√(3)/2 = 14/√(3) ≈ 8.083. No chord may exceed that. The computed AB = 8 clears it by less than a tenth, which is a genuinely tight test: almost any arithmetic slip in the quadratic would have produced a side too long to fit in the circle. The diagonals AC = 7 and BD = 39/7 ≈ 5.571 fit comfortably. Triangle inequalities hold too: △ ACD has sides 3, 5, 7 with 3 + 5 = 8 ＞ 7, and △ ABC has sides 8, 3, 7 with 3 + 7 = 10 ＞ 8. Second, an explicit construction. Put the centre at the origin with R = 7/√(3) ≈ 4.0415 and place A = (4.0415, 0), D = (0.9485, 3.9286), C = (-2.0207, 3.5000), B = (-3.8765, 1.1429), all four at distance R from the origin and in the order A, B, C, D around the circle. Measuring straight off those coordinates: BC = 3.0000, CD = 3.0000, DA = 5.0000, AB = 8.0000, ∠ CDA = 120.00° and ∠ ABC = 60.00°. Every given is reproduced. The diagonals measure AC = 7.0000 and BD = 5.5714, and 5.5714 = 39/7 exactly, so BD is indeed the shorter one. Ptolemy checks as 7 × 5.5714 = 39.000 = 8 · 3 + 3 · 5. Third, a word about the answer list. The longer diagonal 7 does not appear among the five choices, so in a hurry one could have stopped early. That is a fact about how this particular list was written, not a proof, and it would fail on a list that included 7. Meanwhile 39/7 ≈ 5.57 sits between choice (C) 5 and choice (E) 41/7 ≈ 5.86, all of which are perfectly plausible chord lengths for a circle of diameter 8.08, so the list gives away nothing and the explicit comparison in the last step is what actually settles the question.
💡Key takeaway

Cut a cyclic quadrilateral along the diagonal that sits between the two sides you already know, then let the rule that opposite angles add to 180° carry what you learned across to the other side.

  • Draw it and choose the cut
  • Law of Cosines gives AC = 7
  • The circle fixes the angle at B
  • Solve a quadratic for AB
  • Ptolemy delivers the second diagonal
  • Compare, do not assume

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