Question

Triangle $ABC$ in the figure has area $10.$ Points $D,$ $E$ and $F,$ all distinct from $A,$ $B$ and $C,$ are on sides $AB,$ $BC$ and $CA$ respectively, and $AD=2,$ $DB=3.$ If triangle $ABE$ and quadrilateral $DBEF$ have equal areas, then that area is

$4$

$5$

$6$

$\frac53\sqrt{10}$

not uniquely determined

Answer: C

Concepts:triangle area area decomposition similarity

Difficulty rating: 2340

Small Hint:

Draw $DE$ and cancel the common triangle $DBE$ from the equal areas

Big Hint:

Equal areas of $ADE$ and $FDE$ imply $AF\parallel DE$ ; then compare triangles $DBE$ and $ABC$

Solution:

Draw $DE.$ From $\begin{aligned} [ABE]&=[ADE]+[DBE],\\ [DBEF]&=[FDE]+[DBE], \end{aligned}$ equality implies $[ADE]=[FDE].$ Their common base $DE$ then gives $AF\parallel DE,$ so triangles $DBE$ and $ABC$ are similar. Hence $\frac{BE}{BC}=\frac{DB}{AB}=\frac{3}{5}.$ Triangles $ABE$ and $ABC$ share the altitude from $A$ to $BC,$ so $[ABE]=(\frac{3}{5})[ABC]=6.$

Therefore, the correct answer is C.

1983 AMC 12 Problem 28

Answer: C

Difficulty rating: 2340

Small Hint:

Big Hint:

Solution:

Problem 28 in Other Years