AMC 10 Step by Step

2006 AMC 10B Problem 23

P21-25~7 minPrint

A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?

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Key insight

Draw a line from the top corner to where the two lines cross. Triangles with the same height have areas in the ratio of their bases.

Solution

1. Name the points. Call the big triangle ABCABC: AA bottom left, BB bottom right, CC at the top. One line goes from AA to DD on side BCBC; the other goes from BB to EE on side ACAC. They cross at FF. The given areas are: triangle AFE=3AFE = 3, triangle AFB=7AFB = 7, triangle BFD=7BFD = 7. The shaded quadrilateral is CEFDCEFD.
2. Split the quadrilateral. Draw CFCF. It cuts the quadrilateral into two triangles. Call their areas xx (triangle CFECFE) and yy (triangle CFDCFD). The answer is x+yx + 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 ADAD. Triangles AFBAFB and DFBDFB both have top point BB and bases AFAF and FDFD on the same line. Their areas are both 77, so AF=FDAF = FD.
5. Now look at triangles AFCAFC and DFCDFC. Both have top point CC and the equal bases AFAF and FDFD, so their areas are equal:
3+x=y.3 + x = y.
6. Along the line BEBE. Triangles EFAEFA and BFABFA both have top point AA, with bases EFEF and FBFB. Their areas are 33 and 77, so EF:FB=3:7EF : FB = 3 : 7.
7. Now look at triangles EFCEFC and BFCBFC, both with top point CC and bases EFEF and FBFB. Their areas are xx and y+7y + 7, so
xy+7=37,which means7x=3(y+7).\frac{x}{y + 7} = \frac{3}{7}, \quad\text{which means}\quad 7x = 3(y + 7).
8. Solve. Put y=x+3y = x + 3 into the second equation: 7x=3(x+10)=3x+307x = 3(x + 10) = 3x + 30, so 4x=304x = 30 and x=152x = \frac{15}{2}. Then y=152+3=212y = \frac{15}{2} + 3 = \frac{21}{2}.
9. The quadrilateral has area x+y=152+212=362=18x + y = \frac{15}{2} + \frac{21}{2} = \frac{36}{2} = 18.

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Label the triangle ABCABC with AA bottom-left, BB bottom-right, CC on top. The cevians are ADAD (DD on BCBC) and BEBE (EE on ACAC), meeting at FF. The given areas are [AFE]=3[AFE] = 3, [AFB]=7[AFB] = 7, [BFD]=7[BFD] = 7; the shaded quadrilateral is CDFECDFE.

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

Along ADAD: [AFB]=[DFB]=7[AFB] = [DFB] = 7 and both have apex BB, so AF=FDAF = FD. Then [AFC][AFC] and [DFC][DFC] (apex CC, equal bases) are equal:
3+x=y.3 + x = y.

Along BEBE: [EFA]:[BFA]=3:7[EFA] : [BFA] = 3 : 7 with apex AA, so EF:FB=3:7EF : FB = 3 : 7. Then with apex CC,
[EFC]:[BFC]=3:7⟹xy+7=37⟹7x=3y+21.[EFC] : [BFC] = 3 : 7 \quad\Longrightarrow\quad \frac{x}{y + 7} = \frac{3}{7} \quad\Longrightarrow\quad 7x = 3y + 21.

Substitute y=x+3y = x + 3: 7x=3x+307x = 3x + 30, so x=152x = \frac{15}{2} and y=212y = \frac{21}{2}. The quadrilateral has area x+y=18x + y = 18.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.