2018 AMC 10A Problem 24

P21-25~6 minPrint

Triangle Centers & Cevians Similar & Congruent Triangles Set up the equation/formula and compute; no special trick needed Cut the figure into known shapes (triangles, rectangles, sectors)

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$ ?

Show solution without answering

Key insight

Since AB is 5 times AC, the bisector splits both triangle ABC and half-size triangle ADE into areas 5 to 1.

Solution

The quadrilateral $FDBG$ is triangle $ABG$ with triangle $ADF$ cut off the top. So find those two areas.

1. How the bisector splits area. The point $G$ is on the bisector of angle $A$ , so it is the same distance from line $AB$ as from line $AC$ . Call that distance $h$ . Then
$[ABG] = \tfrac12 \cdot AB \cdot h, \qquad [ACG] = \tfrac12 \cdot AC \cdot h.$
Their ratio is $AB : AC = 50 : 10 = 5 : 1$ .
2. Area of $ABG$ . Triangle $ABC$ (area $120$ ) is split into $6$ equal shares, $5$ for $ABG$ and $1$ for $ACG$ :
$[ABG] = \tfrac56 \cdot 120 = 100.$
3. Area of $ADE$ . $D$ and $E$ are midpoints, so $AD = 25$ and $AE = 5$ . Triangle $ADE$ is triangle $ABC$ shrunk to half size, and halving every length divides the area by $2 \cdot 2 = 4$ . So $[ADE] = \tfrac{120}{4} = 30$ .
4. Area of $ADF$ . The same bisector passes through $F$ on $DE$ . By the same reasoning as in step 1, it splits triangle $ADE$ in the ratio $AD : AE = 25 : 5 = 5 : 1$ . So
$[ADF] = \tfrac56 \cdot 30 = 25.$
5. Subtract. $[FDBG] = [ABG] - [ADF] = 100 - 25 = 75$ .

The answer is $\boxed{\textbf{(D)}\ 75}$ .

Why this works

Any point on an angle bisector is equally far from the two sides of the angle, so the two triangles it creates share a height and their areas follow the ratio of the sides. The midpoints make a half-size copy of the whole picture with the same bisector, so the same $5:1$ split repeats inside it. The actual lengths and angles never need to be found.

Alternative approach

Look only at the bottom part. The trapezoid $DBCE$ has area $120 - 30 = 90$ . The bisector splits the whole triangle $5 : 1$ and the top triangle $5 : 1$ , so it also splits what is left, the trapezoid, $5 : 1$ . The part on the $B$ side is $FDBG$ , with area $\tfrac56 \cdot 90 = 75$ .

Concise solution

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

Quadrilateral $FDBG$ is what is left of $\triangle ABG$ after removing $\triangle ADF$ , so compute those two areas.

Triangle $ABG$ . By the angle bisector theorem, $\dfrac{BG}{GC} = \dfrac{AB}{AC} = \dfrac{50}{10} = 5$ , so $BG = \tfrac56 BC$ . Triangles $ABG$ and $ABC$ share the altitude from $A$ , so
$[ABG] = \frac56 [ABC] = \frac56 \cdot 120 = 100.$

Triangle $ADF$ . Since $D$ and $E$ are midpoints, $\triangle ADE$ is $\triangle ABC$ scaled by $\tfrac12$ from $A$ , with area $\tfrac14 \cdot 120 = 30$ . The same bisector from $A$ cuts $DE$ at $F$ , and in $\triangle ADE$ the bisector theorem gives $\dfrac{DF}{FE} = \dfrac{AD}{AE} = \dfrac{25}{5} = 5$ , so
$[ADF] = \frac56 [ADE] = \frac56 \cdot 30 = 25.$

Therefore $[FDBG] = [ABG] - [ADF] = 100 - 25 = 75$ .

The answer is $\boxed{\textbf{(D)}\ 75}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Geometry, Chapter 6: Right Triangles (§6.5)