AMC 10 · 2020 · #24

Grade 8 geometry-2d

equilateral-trianglerotation-isometrypythagorean-theoremthirty-sixty-ninety-trianglespatial-visualization physical-representationidentify-subproblemssymmetry-argument ↑ Prerequisites: equilateral-trianglepythagorean-theorem

📏 Long solution 💡 3 insights

Problem

Inside an equilateral triangle sits exactly one point whose distances to the three vertices are one, root three, and two. Find the triangle's side length.

Pick an answer.

(A)

$1+\sqrt{2}$

(B)

$\sqrt{7}$

(C)

$\frac{8}{3}$

(D)

$\sqrt{5+\sqrt{5}}$

(E)

$2\sqrt{2}$

How to solve

Strategy Visualize Spatial Relationships

The three known lengths radiate out of P like spokes, so no triangle inequality or Pythagorean check can be applied to them directly. Tool #17 (Visualize Spatial Relationships) is the unlock: rotating the figure 60° about a vertex is a rigid motion the equilateral triangle allows, and it carries one spoke on top of another so that 1, √(3), and 2 end up as the three sides of one triangle. Tool #1 (Draw a Diagram) keeps track of where the rotated copy P' lands. Tool #7 (Identify Subproblems) splits the finish into two small right-triangle computations. Tool #3 (Eliminate Possibilities) discards the second algebraic root using the fact that P is inside the triangle.

1STEP 1

Rotate about one vertex

A sixty-degree turn carries the distances across.

AP' = AP = 1, CP' = BP = √(3)

2STEP 2

Spot the small equilateral triangle

The point and its image form an equilateral triangle.

AP = AP' = 1, ∠ PAP' = 60° ⟹ PP' = 1, ∠ AP'P = 60°

3STEP 3

Spot the right triangle

The three distances form a right triangle.

PP'² + P'C² = 1² + (√(3))² = 4 = PC² ⟹ ∠ PP'C = 90°

4STEP 4

Stack the two angles

Adding them gives the large angle.

∠ AP'C = ∠ AP'P + ∠ PP'C = 60° + 90° = 150°

5STEP 5

One perpendicular finishes it

A perpendicular gives the side squared.

AC² = AH² + CH² = (1/2)² + (3√(3)/2)² = 1/4 + 27/4 = 28/4 = 7

6STEP 6

Rule out the bad root

Only the larger root survives, so root seven.

s² ∈ {1, 7}, s ＞ 2 ⟹ s = √(7) ⟹ (B)

Answer

√(7)

Put s = √(7) back into the original picture and measure the three angles at P; they must close up to a full turn. Law of cosines in △ APB with AB² = 7: cos ∠ APB = (1 + 3 - 7)/(2 · 1 · √(3)) = -√(3)/2, so ∠ APB = 150°. In △ BPC: cos ∠ BPC = (3 + 4 - 7)/(2 · √(3) · 2) = 0, so ∠ BPC = 90°. In △ CPA: cos ∠ CPA = (4 + 1 - 7)/(2 · 2 · 1) = -1/2, so ∠ CPA = 120°. The sum is 150° + 90° + 120° = 360° exactly, which is precisely the condition for P to sit inside the triangle — a strong confirmation. Coordinates agree: with A = (0,0), B = (√(7), 0), C = (√(7)/2, √(21)/2), the point P = (5/2√(7), √(3)/2√(7)) has AP = 1, BP = √(3), CP = 2 and lies inside. Note that the crude size test s ＞ 2 alone cannot pick the answer, since (A) 1+√(2) ≈ 2.414, (C) 8/3 ≈ 2.667, (D) √(5+√(5)) ≈ 2.690, and (E) 2√(2) ≈ 2.828 all exceed 2; the exact angle sum is what pins down √(7).

💡Key takeaway

Rotating the whole picture 60° about one vertex slides the three spokes 1, √(3), 2 into a single triangle, and 1² + (√(3))² = 2² makes that triangle right-angled — one perpendicular then gives s² = 1/4 + 27/4 = 7, so s = √(7), with nothing past Grade 8 geometry.

Rotate the figure about A
△ APP' is equilateral
△ PP'C is a right triangle
Stack the two angles at P'
One perpendicular finishes it
Rule out the second root