AMC 10 · 2018 · #24

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
angle-bisector-theoremarea-trianglessimilar-trianglesratio-proportion identify-subproblemscomplementary-counting ↑ Prerequisites: area-trianglessimilar-triangles
📏 Medium solution 💡 3 insights
Problem
In triangle ABC, side AB=50, side AC=10, and the area is 120. Point D is the midpoint of AB and E is the midpoint of AC, so DE is the segment joining the two midpoints. The bisector of angle A cuts DE at F and cuts BC at G. Find the area of the four-sided region FDBG.

Pick an answer.

(A)
60
(B)
65
(C)
70
(D)
75
(E)
80

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

How to solve
Strategy Identify Subproblems

Tool #7 (Identify Subproblems): the quadrilateral FDBG is awkward to attack head-on, but it is exactly the big triangle ABG with the small corner triangle ADF removed. So the job splits into two easy areas. Tool #1 (Draw a Diagram) pins down where F and G sit. Tool #4 (Introduce a Variable) handles the angle-bisector split of BC with a ratio variable. Tool #16 (Change Focus / Count the Complement) is the finishing move: instead of measuring FDBG directly, subtract the corner triangle from the larger triangle.

1STEP 1

Draw and label the figure

Sketch A on top with long AB, short AC; mark midpoints D, E, draw DE, then the bisector meeting DE at F and BC at G.

AB=50, AC=10, [ABC]=120
2STEP 2

Reframe the quadrilateral

Inside triangle ABG, D sits on AB and F on AG, so cutting away corner triangle ADF leaves FDBG: [FDBG] = [ABG] - [ADF].

[FDBG] = [ABG] - [ADF]
3STEP 3

Find [ABG] with the angle bisector

Angle Bisector Theorem gives BG:GC = 50:10 = 5:1, so BG is 5/6 of BC; equal heights make [ABG] = 5/6·120 = 100.

BG:GC = 50:10 = 5:1 → [ABG] = 5/6· 120 = 100
4STEP 4

Find [ADF] by scaling

DE is the midsegment, so F is the midpoint of AG; triangles ADF and ABG share angle A at ratio 1/2, so [ADF] = (1/2)²·100 = 25.

[ADF] = (1/2)² [ABG] = 1/4· 100 = 25
5STEP 5

Subtract to get the quadrilateral

Cut the corner from the big triangle: [FDBG] = 100 - 25 = 75, which is choice (D).

[FDBG] = 100 - 25 = 75
Answer
75
The answer 75 should be a sensible chunk of the whole triangle's area 120. The region FDBG is the lower part of triangle ABG (area 100), so it must be less than 100 but a clear majority of it, since only the small 1/4 corner is removed. 75 fits exactly, and it stays under the total 120. Also 75 = 3/4 · 100, the expected three-quarters left after cutting a quarter-corner.
💡Key takeaway

To get a weird four-sided region, find a triangle around it and just subtract the corner you do not want.

  • Draw and label the figure
  • Reframe the quadrilateral
  • Find [ABG] with the angle bisector
  • Find [ADF] by scaling
  • Subtract to get the quadrilateral

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