Porisma

Worked solution

Use an AP of sector areas and a GP condition

A radius-22 circle is cut into 1010 unequal sectors with areas in arithmetic progression. The first and smallest has area aa, followed by a+d,a+2d,…a+d,a+2d,\ldots square units. Show that 2a+9d=4π52a+9d=\frac{4\pi}{5}. If the first, fourth and ninth sector areas are consecutive GP terms, show that a=π5a=\frac\pi5. Hence find the area of the largest sector.

[2010/CJC/Promo/7(b)(modified)]

CJC · Sequences and Series · Example 33 (Self-Reading)

Total area condition

Answer

2a+9d=4π52a+9d=\frac{4\pi}{5}.

Equate the ten-term AP sum to the circle area

102(2a+9d)=π(2)2⟹2a+9d=4π5\frac{10}{2}(2a+9d)=\pi(2)^2\quad\Longrightarrow\quad2a+9d=\frac{4\pi}{5}

Ten sectors fill a circle of area pi·2².

How did this part go?

Smallest sector area

Answer

a=π5a=\frac\pi5.

Use the selected positions, not adjacent AP terms

a+3da=a+8da+3d⟹a2+6ad+9d2=a2+8ad\frac{a+3d}{a}=\frac{a+8d}{a+3d}\quad\Longrightarrow\quad a^2+6ad+9d^2=a^2+8ad

The first, fourth and ninth areas are a,a+3d,a+8d. Positive areas allow the ratio equation.

Use unequal sectors to exclude zero difference

9d2−2ad=0⟹d=2a99d^2-2ad=0\quad\Longrightarrow\quad d=\frac{2a}{9}

Since the areas increase from the smallest and are unequal, d>0. The source uses d≠0 to divide the factorisation.

Substitute into the area condition

2a+9(2a9)=4π5⟹4a=4π5⟹a=π52a+9\left(\frac{2a}{9}\right)=\frac{4\pi}{5}\quad\Longrightarrow\quad4a=\frac{4\pi}{5}\quad\Longrightarrow\quad a=\frac\pi5

Keep both requested show-that results.

How did this part go?

Largest sector area

Answer

3π5\frac{3\pi}{5} square units.

Use the tenth area

a+9d=a+2a=π5+2π5=3π5a+9d=a+2a=\frac\pi5+\frac{2\pi}{5}=\frac{3\pi}{5}

There are nine increments after the first sector.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Ten increasing sector areas fill a radius-two circle, labelled a,a+d,…,a+9d. The first, fourth and ninth are a,a+3d,a+8d; the largest is a+9d. The redraw uses calculated angles 18°,22°,…,54°.
Ten increasing sector areas fill a radius-two circle, labelled a,a+d,…,a+9d. The first, fourth and ninth are a,a+3d,a+8d; the largest is a+9d. The redraw uses calculated angles 18°,22°,…,54°.
Ten increasing sector areas fill a radius-two circle, labelled a,a+d,…,a+9d. The first, fourth and ninth are a,a+3d,a+8d; the largest is a+9d. The redraw uses calculated angles 18°,22°,…,54°.

Solution notes

The source wedge picture is schematic: its largest wedge is drawn as a quarter-circle, although the calculated area 3π/5 in a radius-two circle corresponds to 54°. This redraw uses calculated proportions; all source labels and the original figure are preserved in source evidence.