AMC 10 · 2011 · #25

Grade 11 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

inscribed-anglearc-measuresine-area-formula extreme-principleidentify-subproblems ↑ Prerequisites: inscribed-angle

📏 Long solution 💡 4 insights

Problem

Three triangle centres and two vertices make a five-sided figure whose area is to be made largest. Find the angle that does it.

Pick an answer.

(A)

$60^{\circ}$

(B)

$72^{\circ}$

(C)

$75^{\circ}$

(D)

$80^{\circ}$

(E)

$90^{\circ}$

How to solve

Strategy Extreme Principle

The question asks for a maximum, so the Extreme Principle is the destination: find the shape of the winning configuration. Getting there needs setup. Naming the angle at B as the one free variable shows the family is one-dimensional. Computing ∠ BOC, ∠ BIC, ∠ BHC separately is three small subproblems, and they all come out 120^° -- the signal to reorganize the five points as one circle instead of five loose objects. On that fixed circle the pentagon's area turns into a sum of sines of arcs, which is an easier related problem with a clean symmetric answer, and working backwards from the winning arcs recovers the angle.

1STEP 1

Pin the angle at B to a range

The free angle is confined to a narrow range.

B + C = 120^°, AC ≥ AB → B ≥ C → 60^° ≤ B ≤ 90^°

2STEP 2

Incenter and orthocenter see BC at 120 degrees

Two centres see the fixed side at the same angle.

∠ BIC = 180^° - (B+C)/2 = 120^°, ∠ BHC = 180^° - ∠ BAC = 120^°

3STEP 3

The circumcenter matches, so five points are concyclic

The third matches, so all five lie on one circle.

∠ BOC = ∠ BIC = ∠ BHC = 120^°

4STEP 4

That circle never changes

That circle never changes size.

r = 1/(2sin 120^°) = 1/√(3), R = 1/(2sin 60^°) = 1/√(3)

5STEP 5

Locate all five points by arcs

Every point is placed by its own arc.

arc CO = 60^°, arc CI = B, arc CH = 2B - 60^°, arc CB = 120^°

6STEP 6

Only I and H can move

Only two of the five actually move.

[BCO] = √(3)/12 (constant), [BCOIH] = [BCO] + [BOIH]

7STEP 7

Trade polygon area for circular segments

The area becomes a sum of sines of arcs.

[BCOIH] = 1/2r²(sinα + sinβ + sinγ) = (sinα + sinβ + sinγ)/6

8STEP 8

Equal arcs win

That sum is largest when the arcs are equal.

sinα + sinβ = 2sin(α+β)/2cos(α-β)/2 ≤ 2sin(α+β)/2, max(sinα+sinβ+sinγ) = 3sin 20^°

9STEP 9

Work back to the angle at B

Working back gives 80 degrees, choice (D).

t = 20^° → ∠ CBA = 60^° + t = 80^°

Answer

80^°

The relation α = β is not an extra assumption slipped in to make the algebra symmetric -- it drops out of the arc computation, and the unrestricted optimum (20^°, 20^°, 20^°) happens to satisfy it. That matters, because I and H are not free to be dragged around the arc; they are determined by the triangle. Since the free optimum lands inside the one-parameter family the triangle actually traces out, the value found is a true maximum. A direct check backs this up: with t = ∠ CBA - 60^° the area is (2sin t + sin(60^° - 2t))/6, which evaluates to about 0.1443 at 60^°, 0.1673 at 72^°, 0.1696 at 75^°, 0.1710 at 80^°, and 0.1667 at 90^°, so (D) beats every other choice. Both endpoints are degenerate anyway -- at 60^° the triangle is equilateral and O, I, H collapse into one point, and at 90^° the orthocenter falls onto B -- and both give smaller area, so the maximum is comfortably interior.

💡Key takeaway

Because ∠ A is 60^°, the three centers O, I, H all see BC at 120^°, so they land with B and C on one circle that never changes; the pentagon's area is then just 1/6 of the sum of the sines of three arcs that always total 60^°, and that sum is biggest when the three arcs are equal.

Pin the angle at B to a range
Incenter and orthocenter see BC at 120 degrees
The circumcenter matches, so five points are concyclic
That circle never changes
Locate all five points by arcs
Only I and H can move
Trade polygon area for circular segments
Equal arcs win
Work back to the angle at B