AMC 10 · 2015 · #13

Grade 10 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
inscribed-anglecyclic-quadrilateralangle-sum-triangleisosceles-triangle identify-subproblems ↑ Prerequisites: inscribed-angleangle-sum-triangle
📏 Medium solution 💡 3 insights
Problem
Four points on a circle carry two known angles and two known sides. Find a diagonal.

Pick an answer.

(A)
$3+\sqrt{5}$
(B)
6
(C)
$\dfrac{9}{2}\sqrt{2}$
(D)
$8-\sqrt{2}$
(E)
7
How to solve
Strategy Organize Information in More Ways

The four givens arrive filed under four different letters, and as long as they stay that way nothing connects them: the 40° lives at D, the 70° and the wanted length live around A and C, and D is not even a vertex of the triangle that contains AC. The fix is to re-file the data. An inscribed angle does not really belong to its vertex; it belongs to the chord its two rays land on. Sorted that way, ∠ ADB and the unnamed angle ∠ ACB turn out to be filed under the same chord AB, which is exactly the bridge that carries the 40° off of D and into triangle ABC, where the 70° and the length 6 already are. After that the triangle closes itself. Reaching first for the law of cosines or for Ptolemy's theorem is the tempting alternative and it stalls, because both need lengths that the problem never supplies. Two supporting moves finish the job honestly: ruling out the second reading of the inscribed-angle theorem rather than assuming the picture, and a boundary check on the unused length AD to confirm the four givens can actually coexist.

1STEP 1

Read the order out of the name

The name fixes the order around the circle.

order around the circle: A → B → C → D → A; C, D on the same side of chord AB
2STEP 2

Re-file each angle under its chord

Two angles stand on the same chord.

∠ ADB and ∠ ACB both stand on chord AB → ∠ ACB = 40° or ∠ ACB = 140°
3STEP 3

Close the triangle, kill the other case

The angle sum kills the other case.

70° + 140° = 210° ＞ 180° → ∠ ACB = 40°, ∠ ABC = 180° - (70° + 40°) = 70°
4STEP 4

Match equal angles to opposite sides

Two equal angles make a triangle isosceles.

∠ BAC = ∠ ABC = 70° → AC = BC = 6
5STEP 5

Check the unused given is not a lie

So the diagonal is 6, choice (C).

arc BC = 140°, arc AB = 80° → arc CD + arc DA = 140°; 2R = 6/(sin 70°) ≈ 6.385, 0 ＜ AD ＜ 6
Answer
6
Rebuild the figure from scratch and measure. Put the circle at radius R = 3/sin 70° ≈ 3.1925 and place the four points at arc positions 0°, 80°, 220°, and 282.4°. Then BC = 6.000, AD = 4.000, ∠ BAC = 70.0°, ∠ ADB = 40.0° — all four givens reproduced — and AC = 6.000, matching choice (B). The points also come out in the order A, B, C, D with all four arcs positive, so this really is a quadrilateral and not a crossed figure. The result also passes a rough sanity test: ∠ ADB = 40° is smaller than ∠ BAC = 70°, so chord AB should be shorter than chord BC, and indeed AB ≈ 4.10 ＜ 6. Three of the wrong choices cluster just above 6 — (9/2)√(2) ≈ 6.364, 8 - √(2) ≈ 6.586, and 7 — which is what makes an eyeballed diagram useless here, since 2R ≈ 6.385 is the diameter and (9/2)√(2) is close enough to it to look plausible in a sketch; only the exact angle argument separates them. Choice (A) 3+√(5) ≈ 5.24 is the only option below 6, and it can be discarded on sight: AC ＜ BC would force the angle at B to be smaller than the 70° angle at A, hence ∠ ACB ＞ 40°, contradicting the transferred inscribed angle.
💡Key takeaway

An inscribed angle belongs to the chord it looks at, not to its vertex, so the 40° at D is really information about chord AB and can be moved to C, which turns triangle ABC into a 70-70-40 triangle whose two equal angles look out on two equal sides.

  • Read the order out of the name
  • Re-file each angle under its chord
  • Close the triangle, kill the other case
  • Match equal angles to opposite sides
  • Check the unused given is not a lie

More like this

Same archetype — closest grade level first.