2013 AMC 10B Problem 16

P16-20~4 minPrint

Triangle Centers & Cevians Quadrilaterals & Polygon Areas Triangles: Area & Pythagorean Set up the equation/formula and compute; no special trick needed Cut the figure into known shapes (triangles, rectangles, sectors)

In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ , $PE=1.5$ , $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$ ?

Show solution without answering

Key insight

1.5, 2, 2.5 is half of a 3-4-5 triangle, so the medians cross at a right angle. Doubling PD and PE gives the other pieces of each median.

Solution

1. Look at triangle $PED$ first, since we know all three of its sides: $1.5$ , $2$ , $2.5$ . Doubling them gives $3, 4, 5$ , a right triangle. Check: $1.5^2 + 2^2 = 2.25 + 4 = 6.25 = 2.5^2$ . So the angle at $P$ is $90^\circ$ , which means the medians $AD$ and $CE$ cross at right angles.
2. $P$ is where two medians meet, so it is the centroid. The centroid cuts each median into two pieces, and the piece touching the corner is twice as long as the piece touching the midpoint. So
$AP = 2 \cdot PD = 4, \qquad CP = 2 \cdot PE = 3.$
3. The segments $AD$ and $CE$ are the diagonals of $AEDC$ , and they cross at $P$ . They cut $AEDC$ into four right triangles, all with their right angle at $P$ . Each has area half the product of its two legs:

Triangle	Legs	Area
$APE$	$4$ and $1.5$	$3$
$EPD$	$1.5$ and $2$	$1.5$
$DPC$	$2$ and $3$	$3$
$CPA$	$3$ and $4$	$6$

Add them up: $3 + 1.5 + 3 + 6 = 13.5$ .

The answer is $\boxed{\textbf{(B)}\ 13.5}$ .

Why this works

Two facts do all the work. Three given lengths that fit the Pythagorean theorem reveal a hidden right angle, and the centroid's $2$ to $1$ split turns the short pieces into full medians. Whenever the diagonals of a four-sided shape cross at right angles, splitting it into four right triangles makes the area easy.

Alternative approach

A four-sided shape whose diagonals cross at right angles has area equal to half the product of the diagonals (the four right triangles above add up to exactly that). Here $AD = 4 + 2 = 6$ and $CE = 3 + 1.5 = 4.5$ , so the area is $\frac{1}{2} \cdot 6 \cdot 4.5 = 13.5$ .

Concise solution

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

The three given lengths satisfy $1.5^2 + 2^2 = 2.25 + 4 = 6.25 = 2.5^2$ , so triangle $PED$ is right-angled at $P$ . In other words, the medians $AD$ and $CE$ are perpendicular.

$P$ is the centroid, which cuts each median in a $2 : 1$ ratio measured from the vertex. Hence
$AP = 2 \cdot PD = 4, \qquad CP = 2 \cdot PE = 3,$
so $AD = 4 + 2 = 6$ and $CE = 3 + 1.5 = 4.5$ .

Quadrilateral $AEDC$ has diagonals $AD$ and $CE$ , which are perpendicular. Its area is half the product of the diagonals:
$\frac{1}{2} \cdot 6 \cdot 4.5 = 13.5.$

The answer is $\boxed{\textbf{(B)}\ 13.5}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.5)