AMC 10 · 2015 · #19

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
isosceles-trianglethirty-sixty-ninety-trianglearea-triangles convert-to-algebra ↑ Prerequisites: thirty-sixty-ninety-trianglearea-triangles
📏 Medium solution 💡 3 insights
Problem
Triangle ABC is an isosceles right triangle with its right angle at C and area 12.5. Two rays from C cut the right angle into three equal angles; these rays meet the hypotenuse AB at points D and E. Find the area of triangle CDE.

Pick an answer.

(A)
$\dfrac{5\sqrt{2}}{3}$
(B)
$\dfrac{50\sqrt{3}-75}{4}$
(C)
$\dfrac{15\sqrt{3}}{8}$
(D)
$\dfrac{50-25\sqrt{3}}{2}$
(E)
$\dfrac{25}{6}$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The whole problem lives in one picture, so Tool #1 (Draw a Diagram) is the spine: drawing the triangle and the two trisecting rays exposes the 30° angles and the symmetry that makes CD=CE. The middle triangle CDE is awkward to attack head-on, so Tool #7 (Identify Subproblems) reframes it: drop a perpendicular and the figure breaks into a 45-45-90 triangle and a 30-60-90 triangle whose side ratios are known, and then CDE is just the whole triangle minus the two equal corner triangles. Tool #4 (Introduce a Variable) names that perpendicular height so the two known ratios meet in a single equation that pins it down.

1STEP 1

Find the leg length

The area of a right triangle is half the product of its legs; solving 12\frac{1}{2} a² = 12.5 gives a = 5.

12\frac{1}{2} a² = 12.5 → a² = 25 → a = 5
2STEP 2

Mark the trisected angles

Trisecting the 90° angle at C gives ∠ ACD=∠ DCE=∠ ECB=30°; base angles are 45° each.

∠ ACD=∠ DCE=∠ ECB=30°, ∠ CAB=∠ CBA=45°
3STEP 3

Drop a perpendicular to make special triangles

Drop a perpendicular from D to CA at F: this splits the corner into a 30-60-90 triangle (∠ DCF=30°) and a 45-45-90 triangle (∠ DAF=45°).

CF=√3 DF, CD=2 DF, AF=DF
4STEP 4

Solve for the height and the corner area

Setting CF + AF = 5 pins down h = 5(3−1)2\frac{5(\sqrt{3}-1)}{2}, so triangle ACD's area is 25(3−1)4\frac{25(\sqrt{3}-1)}{4}.

h(√3+1)=5→ h=5(3−1)2\frac{5(\sqrt{3}-1)}{2}, [ACD]=12\frac{1}{2}·5· h=25(3−1)4\frac{25(\sqrt{3}-1)}{4}
5STEP 5

Subtract the two equal corners

By symmetry BCE≅ACD, so [CDE] = [ABC] - 2 [ACD] = 50−2532\frac{50 - 25\sqrt{3}}{2}, choice (D).

[CDE]=12.5-2·25(3−1)4\frac{25(\sqrt{3}-1)}{4}=50−2532\frac{50-25\sqrt{3}}{2} (D)
Answer
50−2532\frac{50-25\sqrt{3}}{2}
Estimate sizes: √3≈1.732, so [CDE]≈50-43.32\frac{43.3}{2}≈3.35. That is a small slice of the total area 12.5, which fits a 30° wedge squeezed between two wider 30° corners that each reach toward a 45° base angle, so the middle piece should indeed be the smallest of the three. This rules out the larger-looking options and matches (D), whose value is about 3.35. The two equal corner triangles take up 12.5-3.35≈9.15, a sensible majority of the figure.
💡Key takeaway

Trisecting the right angle makes 30° wedges; drop a perpendicular to split off known triangles, then subtract the two matching corners.

  • Find the leg length
  • Mark the trisected angles
  • Drop a perpendicular to make special triangles
  • Solve for the height and the corner area
  • Subtract the two equal corners

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