AMC 10 · 2006 · #23

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglesratio-proportionsystems-of-equations convert-to-algebra ↑ Prerequisites: area-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A triangle is cut by two cevians, each running from a vertex to a point on the opposite side, one drawn from each of two vertices. The two cevians cross at a point inside the triangle, splitting it into three smaller triangles and one quadrilateral. The two triangles lying on either side of that crossing point along one cevian each have area 7, and the third triangle has area 3. Find the area of the quadrilateral piece.

Pick an answer.

(A)

(B)

(C)

$\frac{35}{2}$

(D)

(E)

$\frac{55}{3}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The figure is one compound shape, but every piece is a triangle sharing a side (a cevian) with its neighbors. The winning move is to break the figure into triangles that share the same height: for those, area ratio equals base ratio, so each given area quietly reports a length ratio. A clean labeled diagram fixes which region is which, and naming the one still-unknown sub-triangle turns those ratios into a single equation to solve.

1STEP 1

Label the points and pieces

Name the triangle ABC: one cevian runs from A to D on BC, the other from B to E on AC, and they cross at F. The shaded piece is FDCE.

Pieces: [AFE]=3, [ABF]=7, [BFD]=7, [FDCE]=?

2STEP 2

Equal areas force a midpoint

Triangles ABF and BFD share the height from B and have equal areas 7 and 7, so their bases match: AF = FD, making F the midpoint of AD.

[ABF]/[BFD]=AF/FD=7/7=1 → AF=FD

3STEP 3

Read the split on the other cevian

Same rule on cevian BE: ABF and AFE share the height from A with bases BF and FE, so BF : FE = 7 : 3.

BF/FE=[ABF]/[AFE]=7/3

4STEP 4

Cut the quadrilateral and name a variable

Draw CF, splitting FDCE into CFD and CFE; let [CFE] = x. The midpoint gives [CFD] = [CFA] = 3 + x, so [FDCE] = 3 + 2x.

[CFD]=[CFA]=[AFE]+[CFE]=3+x, [FDCE]=(3+x)+x=3+2x

5STEP 5

Use the 7:3 split to solve for x

BFC and EFC share apex C with bases BF and FE, so they hold the same 7 : 3 ratio: (10 + x) : x = 7 : 3, giving x = $\frac{15}{2}$ .

(10+x)/x=7/3 → 3(10+x)=7x → 30=4x → x=15/2

6STEP 6

Add up the quadrilateral

Substituting gives [FDCE] = 3 + 2x = 3 + 15 = 18, so the shaded quadrilateral has area 18 — choice (D).

[FDCE]=3+2x=3+2·15/2=3+15=18

Answer

Check the pieces sum sensibly and re-derive from the two cevians. The four areas total 3 + 7 + 7 + 18 = 35. Splitting by cevian AD: [ABD] = 7 + 7 = 14 and [ACD] = 3 + 18 = 21, so BD : DC = 14 : 21 = 2 : 3. Splitting by cevian BE: [ABE] = 7 + 3 = 10 and [CBE] = 7 + 18 = 25, so AE : EC = 10 : 25 = 2 : 5. Both side-ratios are clean, and each split re-gives the quadrilateral as 21 - 3 = 18 and 25 - 7 = 18. Everything agrees, and 18 is choice (D).

💡Key takeaway

When two triangles share the same height, their areas line up exactly with their bases, so every area label is secretly telling you a length ratio.

Label the points and pieces
Equal areas force a midpoint
Read the split on the other cevian
Cut the quadrilateral and name a variable
Use the 7:3 split to solve for x
Add up the quadrilateral