AMC 10 · 2017 · #24

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
equilateral-trianglecoordinate-geometrycentroid-2-to-1symmetry-argument symmetry-argumentidentify-subproblems ↑ Prerequisites: coordinate-geometry
📏 Long solution 💡 3 insights
Problem
An equilateral triangle has all three corners sitting on the curve xy = 1. A vertex of that curve (one of the two turning points closest to the origin) is the centroid, the balance point, of the triangle. Find the square of the triangle's area.

Pick an answer.

(A)
48
(B)
60
(C)
108
(D)
120
(E)
169

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Graph the curve to see where its vertices are and how the whole picture is mirror-symmetric about the line y = x. That symmetry pins down one corner of the triangle for free. From there the work splits into clean subproblems: measure the centroid-to-corner distance (the circumradius), turn that radius into the side length using the fixed shape of an equilateral triangle, and finally read off the area.

1STEP 1

Find the hyperbola's vertices

The curve xy = 1 is symmetric about y = x; its turning points nearest the origin are (1, 1) and (-1, -1). Take the centroid G = (-1, -1).

xy=1 → vertices (1,1) and (-1,-1); G=(-1,-1)
2STEP 2

Use symmetry to place one corner

The curve and centroid are both symmetric across y = x, so the triangle is too — forcing one vertex onto that line: A = (1, 1).

reflect across y=x → A=(1,1)
3STEP 3

Measure the circumradius

Centroid = circumcenter, so R is the distance from G = (-1, -1) to A = (1, 1): R = √8, giving R² = 8.

R=√((1-(-1))²+(1-(-1))²)=√(4+4)=√(8), R²=8
4STEP 4

Turn the radius into the side length

For an equilateral triangle R = s/√3, so squaring gives s² = 3R² = 24.

R=s/√(3) → s²=3R²=3· 8=24
5STEP 5

Square the area

Area = (√3/4)s² = 6√3, so the area squared is (6√3)² = 108 — choice (C).

[△]=√(3)/4s²=√(3)/4· 24=6√(3); [△]²=(6√(3))²=108
Answer
108
Check that the other two corners really are on the curve. The circle (x+1)² + (y+1)² = 8 meets xy = 1 where x + 1/x = 2 (giving (1,1)) or x + 1/x = -4 (giving x = -2 ± sqrt(3)). Those two extra points are (-2+sqrt(3), -2-sqrt(3)) and (-2-sqrt(3), -2+sqrt(3)), both on xy = 1. Their average with (1,1) is (-1,-1) = G, confirming the centroid, and the distance between them is sqrt(24) = 2sqrt(6) = s, matching. An area of 6sqrt(3) is about 10.4, so its square near 108 lands squarely on choice (C).
💡Key takeaway

When a shape is symmetric, the balance point is also the center of its circle, so one corner-to-center distance unlocks everything else.

  • Find the hyperbola's vertices
  • Use symmetry to place one corner
  • Measure the circumradius
  • Turn the radius into the side length
  • Square the area

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