Geometry & Figures

Depth · one concept from first principles · step 2 of 4

Bisect a pentagon by a line from one vertex

As shown in the figure, pentagon ABCDE\text{ABCDE} has vertices A(1,4)\text{A}(1, 4), B(−2,2)\text{B}(-2, 2), C(−1,0)\text{C}(-1, 0), D(3,0)\text{D}(3, 0), and E(3,3)\text{E}(3, 3). A line through point A\text{A} divides the area of pentagon ABCDE\text{ABCDE} into two equal parts. Find the equation of this line.

xy A B C O D E
FunctionsGeometry
Your answer
How to solve
Strategy Identify Subproblems — Split the pentagon with the vertical line x=1 through A into two pieces whose areas are easy to find. They come out almost equal, so the bisector only needs to tilt a little from vertical, crossing side CD at an unknown point M. A thin triangle makes up the difference, and its area fixes M.
1STEP 1

Cut along x = 1

Drop P(1, 0) below A and Q(-2, 0) below B. The right piece APDE is a trapezoid with sides 4 and 3, width 2.

1/2×(4+3) × 2=7
2STEP 2

Area of the left piece

The left piece ABCP is trapezoid ABQP (sides BQ=2, AP=4, width 3) minus triangle BQC (legs 2 and 1).

1/2×(2+4) × 3-1/2 × 2 × 1=9-1=8
3STEP 3

Shift area with a thin triangle

Half the area is 15/2, so the left piece (8) is 1/2 too big. Tilt the cut to M(t, 0), handing triangle AMP (height 4, base 1-t) right.

1/2×(1-t) × 4=1/2, 1-t=1/4, t=3/4
4STEP 4

Equation of line AM

The line passes through A(1, 4) and M(3/4, 0).

(4-0)/(1-3/4)=16, 4=16 × 1+b, b=-12
Answer
y=16x-12
The line y=16x-12 meets y=0 at x=3/4, which is on side CD (between -1 and 3). Right piece AMDE by the shoelace formula with A(1, 4), M(3/4, 0), D(3, 0), E(3, 3): 1/2×(18-4×3/4)=15/2, exactly half of 15.
Takeaway

Cut where the areas are easy, then slide the cut by just the leftover amount.

  • Cut along x = 1
  • Area of the left piece
  • Shift area with a thin triangle
  • Equation of line AM
Where next?
Another one like thissuggested
Same look, different logicsuggested

A parent dashboard for the family lives at sensimlab.com.

▶ Practice — 12 problems