2006 AMC 10B Problem 23

1. Name the points. Call the big triangle $ABC$: $A$ bottom left, $B$ bottom right, $C$ at the top. One line goes from $A$ to $D$ on side $BC$; the other goes from $B$ to $E$ on side $AC$. They cross at $F$. The given areas are: triangle $AFE = 3$, triangle $AFB = 7$, triangle $BFD = 7$. The shaded quadrilateral is $CEFD$.
2. Split the quadrilateral. Draw $CF$. It cuts the quadrilateral into two triangles. Call their areas $x$ (triangle $CFE$) and $y$ (triangle $CFD$). The answer is $x + y$.
3. The tool. If two triangles have the same height (same top point, bases on the same line), their areas compare the same way their bases do. For example, if one base is twice as long, that area is twice as big.
4. Along the line $AD$. Triangles $AFB$ and $DFB$ both have top point $B$ and bases $AF$ and $FD$ on the same line. Their areas are both $7$, so $AF = FD$.
5. Now look at triangles $AFC$ and $DFC$. Both have top point $C$ and the equal bases $AF$ and $FD$, so their areas are equal:
$$3 + x = y.$$
6. Along the line $BE$. Triangles $EFA$ and $BFA$ both have top point $A$, with bases $EF$ and $FB$. Their areas are $3$ and $7$, so $EF : FB = 3 : 7$.
7. Now look at triangles $EFC$ and $BFC$, both with top point $C$ and bases $EF$ and $FB$. Their areas are $x$ and $y + 7$, so
$$\frac{x}{y + 7} = \frac{3}{7}, \quad\text{which means}\quad 7x = 3(y + 7).$$
8. Solve. Put $y = x + 3$ into the second equation: $7x = 3(x + 10) = 3x + 30$, so $4x = 30$ and $x = \frac{15}{2}$. Then $y = \frac{15}{2} + 3 = \frac{21}{2}$.
9. The quadrilateral has area $x + y = \frac{15}{2} + \frac{21}{2} = \frac{36}{2} = 18$.

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

When lines from the corners cut a triangle into pieces with known areas, each crossing point splits a line in some ratio. That ratio shows up in the triangles on both sides of the line, once with known areas and once with unknown ones. Drawing the extra segment from the third corner turns the quadrilateral into two triangles, so you get exactly two equations for the two unknowns.

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 $\boxed{\textbf{(D)}\ 18}$.