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(12\frac{1}{2},√32\frac{√3}{2}), C(-12\frac{1}{2},√32\frac{√3}{2}), D(-1,0), E(-12\frac{1}{2},-√32\frac{√3}{2}), F(12\frac{1}{2},-√32\frac{√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(34\frac{3}{4},√34\frac{√3}{4}), Y(-34\frac{3}{4},√34\frac{√3}{4}), Z(0,-√32\frac{√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√34\frac{3√3}{4}.

[△ ACE]=√34\frac{√3}{4}(√3)²=3√34\frac{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√332\frac{3√3}{32}.

XZ∩ AC=(58\frac{5}{8},√38\frac{√3}{8}), XZ∩ AE=(14\frac{1}{4},-√34\frac{√3}{4})→[corner]=3√332\frac{3√3}{32}
5STEP 5

Add up the answer

Subtract the three equal corners from △ACE: over 32, that's 24√332\frac{24√3}{32} − 9√332\frac{9√3}{32} = 15√332\frac{15√3}{32} — choice (C).

3√34\frac{3√3}{4}-3·3√332\frac{3√3}{32}=24√332\frac{24√3}{32}-9√332\frac{9√3}{32}=15√332\frac{15√3}{32} → (C)
Answer
1532\frac{15}{32}√(3)
The overlap must be smaller than each triangle it sits inside. Here [△ ACE]=3√34\frac{3√3}{4}=24√332\frac{24√3}{32} and [△ XYZ]=√34\frac{√3}{4}(32\frac{3}{2})²=9√316\frac{9√3}{16}=18√332\frac{18√3}{32}, and our answer 15√332\frac{15√3}{32} is below both — exactly what a true intersection should do. Writing the choices over 32 gives A=12√332\frac{12√3}{32}, B=14√332\frac{14√3}{32}, C=15√332\frac{15√3}{32}, D=16√332\frac{16√3}{32}, E=18√332\frac{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=12\frac{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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