AMC 10 · 2018 · #24

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
area-trianglesequilateral-trianglecoordinate-geometrysymmetry-argument identify-subproblemssymmetry-argumentarea-difference ↑ Prerequisites: area-triangles
📏 Medium solution 💡 3 insights
Problem
A regular hexagon ABCDEF has side length 1. Points X, Y, Z are the midpoints of sides AB, CD, EF. Triangle ACE (joining alternate corners) and triangle XYZ (joining the three midpoints) overlap, and their overlap is a convex hexagon. Find the area of that overlap.

Pick an answer.

(A)
$\frac {3}{8}\sqrt{3}$
(B)
$\frac {7}{16}\sqrt{3}$
(C)
$\frac {15}{32}\sqrt{3}$
(D)
$\frac {1}{2}\sqrt{3}$
(E)
$\frac {9}{16}\sqrt{3}$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram): dropping the figure onto a coordinate grid turns "which points are inside both triangles" into exact equations of lines, so the fuzzy picture becomes numbers. Tool #4 (Introduce a Variable): coordinates let me write each side of each triangle as a line equation, and crossing points come from solving those equations. Tool #7 (Identify Subproblems): the hexagon has 120° rotational symmetry, so the three corners that get sliced off △ ACE are identical — I only have to measure one. Tool #7 (Identify Subproblems): instead of the six-sided overlap directly, I compute the big triangle's area and subtract the three equal corners, two easy pieces instead of one hard one.

1STEP 1

Pin the hexagon to a grid

Put center O at the origin; a regular hexagon's side equals its radius, so the six corners sit on a circle of radius 1.

A(1,0), B(1/2,√3/2), C(-1/2,√3/2), D(-1,0), E(-1/2,-√3/2), F(1/2,-√3/2)
2STEP 2

Find the midpoints; spot two equal triangles

Averaging endpoints gives the midpoints; AC = √3, so △ACE is equilateral and △XYZ is too — both centered at O, turned 30° apart.

X(3/4,√3/4), Y(-3/4,√3/4), Z(0,-√3/2); AC=√3
3STEP 3

Cut the overlap out of triangle ACE

Inside △ACE, each side of △XYZ slices off one equal corner near A, C, E, so overlap = △ACE − 3 corners; △ACE has area 3√3/4.

[△ ACE]=√3/4(√3)²=3√3/4; overlap=[△ ACE]-3·[corner]
4STEP 4

Measure one corner triangle

At corner A, sides AC, AE of △ACE and side XZ of △XYZ bound a small triangle; the shoelace formula gives its area 3√3/32.

XZ∩ AC=(5/8,√3/8), XZ∩ AE=(1/4,-√3/4)→[corner]=3√3/32
5STEP 5

Add up the answer

Subtract the three equal corners from △ACE: over 32, that's 24√3/32 − 9√3/32 = 15√3/32 — choice (C).

3√3/4-3·3√3/32=24√3/32-9√3/32=15√3/32 → (C)
Answer
15/32√(3)
The overlap must be smaller than each triangle it sits inside. Here [△ ACE]=3√3/4=24√3/32 and [△ XYZ]=√3/4(3/2)²=9√3/16=18√3/32, and our answer 15√3/32 is below both — exactly what a true intersection should do. Writing the choices over 32 gives A=12√3/32, B=14√3/32, C=15√3/32, D=16√3/32, E=18√3/32. Notice E equals [△ XYZ] exactly: that is the trap for anyone who assumes △ XYZ lies fully inside △ ACE — it does not, since its corners poke out. C sits just under D=1/2√3, matching a hexagon that fills most of △ XYZ.
💡Key takeaway

Drop the figure onto coordinates, then build the overlap by cutting three equal corners off the bigger triangle.

  • Pin the hexagon to a grid
  • Find the midpoints; spot two equal triangles
  • Cut the overlap out of triangle ACE
  • Measure one corner triangle
  • Add up the answer

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