2006 AMC 10B Problem 23

Label the triangle $ABC$ with $A$ bottom-left, $B$ bottom-right, $C$ on top. The cevians are $AD$ ($D$ on $BC$) and $BE$ ($E$ on $AC$), meeting at $F$. The given areas are $[AFE] = 3$, $[AFB] = 7$, $[BFD] = 7$; the shaded quadrilateral is $CDFE$.

Draw $CF$, cutting the quadrilateral into $[CFE] = x$ and $[CFD] = y$. Use one fact repeatedly: two triangles with the same altitude have areas proportional to their bases.

Along $AD$: $[AFB] = [DFB] = 7$ and both have apex $B$, so $AF = FD$. Then $[AFC]$ and $[DFC]$ (apex $C$, equal bases) are equal:
$$3 + x = y.$$

Along $BE$: $[EFA] : [BFA] = 3 : 7$ with apex $A$, so $EF : FB = 3 : 7$. Then with apex $C$,
$$[EFC] : [BFC] = 3 : 7 \quad\Longrightarrow\quad \frac{x}{y + 7} = \frac{3}{7} \quad\Longrightarrow\quad 7x = 3y + 21.$$

Substitute $y = x + 3$: $7x = 3x + 30$, so $x = \frac{15}{2}$ and $y = \frac{21}{2}$. The quadrilateral has area $x + y = 18$.

The answer is $\boxed{\textbf{(D)}\ 18}$.

When cevians cut a triangle into pieces of known area, every ratio of segments along a cevian shows up twice: as a ratio of areas of triangles on one side of it and again on the other side. Drawing the extra segment from the third vertex to the intersection converts the quadrilateral into two triangles that participate in those ratios, yielding exactly as many equations as unknowns.

Mass points: $[AFB] = [DFB]$ gives $AF = FD$, so put mass $1$ at $A$ and $D$; $EF : FB = 3 : 7$ gives masses $7$ at $E$ and $3$ at $B$ after scaling. Balancing at $D$ and $E$ yields $CE : EA$ and $CD : DB$, from which $[ABC] = 35$ and the quadrilateral is $35 - 17 = 18$.

• Assuming the two lines are medians or that the quadrilateral is a parallelogram, or guessing the answer from 3 + 7 + 7 and a total of 34 or 35.
• Setting up the ratio $x : (y + 7)$ as $x : y$ (forgetting that triangle $BFC$ includes the piece of area $7$).