1983 AMC 12 Problem 28

Attempt Problem 28 of the 1983 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1983 AMC 12 solutions, or check the answer key.

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28.

Triangle ABCABC in the figure has area 10.10. Points D,D, EE and F,F, all distinct from A,A, BB and C,C, are on sides AB,AB, BCBC and CACA respectively, and AD=2,AD=2, DB=3.DB=3. If triangle ABEABE and quadrilateral DBEFDBEF have equal areas, then that area is
44
55
66
5310\frac53\sqrt{10}
not uniquely determined
Answer: C
Concepts:triangle areaarea decompositionsimilarity
Small hint:
Draw DEDE and cancel the common triangle DBEDBE from the equal areas
Big hint:
Equal areas of ADEADE and FDEFDE imply AF∥DEAF\parallel DE; then compare triangles DBEDBE and ABCABC
Solution:
Draw DE.DE. From [ABE]=[ADE]+[DBE],[DBEF]=[FDE]+[DBE], \begin{aligned} [ABE]&=[ADE]+[DBE],\\ [DBEF]&=[FDE]+[DBE], \end{aligned} equality implies [ADE]=[FDE].[ADE]=[FDE]. Their common base DEDE then gives AF∥DE,AF\parallel DE, so triangles DBEDBE and ABCABC are similar. Hence BEBC=DBAB=35.\frac{BE}{BC}=\frac{DB}{AB}=\frac{3}{5}. Triangles ABEABE and ABCABC share the altitude from AA to BC,BC, so [ABE]=(35)[ABC]=6.[ABE]=(\frac{3}{5})[ABC]=6.
Therefore, the correct answer is C.
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