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# Circumscribed circle of the hexagon | PRMO 2018 | Problem 7

Try this beautiful problem from Geometry based on Radius of the circumscribed circle of the hexagon

## Hexagon|PRMO| 2007

A point P in the interior of a regular hexagon is at distance 8,8,16 units from three consecutive vertices of the hexagon, respectively. If r is radius of the circumscribed circle of the hexagon, what is the integer closest to r ?

• $12$
• $14$
• $16$

### Key Concepts

Geometry

Triangle

Hexagon

Answer:$14$

PRMO (2018) Problem 7

Pre College Mathematics

## Try with Hints

Show $$\triangle PAB$$is similar to $$\triangle PFC$$.

Can you now finish the problem ..........

The circumradius of a regular hexagon = side of regular hexagon

can you finish the problem........

Note that CF = 2AB, PA = 2PC & PB = 2PF PFC, hence $$\triangle PAB$$is similar to $$\triangle PFC$$.Hence $$A_1P_1C and B_1P_1F$$  are collinear. Let each side of hexagon be equal to x . Let Q & R be foot of altitudes from P to base AB & CF respectively. So R is centre of hexagon

Now

$$\frac{1}{3} \times \frac {\sqrt 3 x}{2}$$ =$$\sqrt{64-\frac{x^2}{4}}$$

$$\Rightarrow 64-\frac{x^2}{4}$$

$$\Rightarrow \frac{4 x^2}{12} =64$$

$$\Rightarrow x=8\sqrt 3$$

We know that the circumradius of a regular hexagon = side of regular hexagon

Hence r=8$$\sqrt 3$$$$\approx$$ 13.856

Therefore r=14 ( nearest integer)

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Try this beautiful problem from Geometry based on Radius of the circumscribed circle of the hexagon

## Hexagon|PRMO| 2007

A point P in the interior of a regular hexagon is at distance 8,8,16 units from three consecutive vertices of the hexagon, respectively. If r is radius of the circumscribed circle of the hexagon, what is the integer closest to r ?

• $12$
• $14$
• $16$

### Key Concepts

Geometry

Triangle

Hexagon

Answer:$14$

PRMO (2018) Problem 7

Pre College Mathematics

## Try with Hints

Show $$\triangle PAB$$is similar to $$\triangle PFC$$.

Can you now finish the problem ..........

The circumradius of a regular hexagon = side of regular hexagon

can you finish the problem........

Note that CF = 2AB, PA = 2PC & PB = 2PF PFC, hence $$\triangle PAB$$is similar to $$\triangle PFC$$.Hence $$A_1P_1C and B_1P_1F$$  are collinear. Let each side of hexagon be equal to x . Let Q & R be foot of altitudes from P to base AB & CF respectively. So R is centre of hexagon

Now

$$\frac{1}{3} \times \frac {\sqrt 3 x}{2}$$ =$$\sqrt{64-\frac{x^2}{4}}$$

$$\Rightarrow 64-\frac{x^2}{4}$$

$$\Rightarrow \frac{4 x^2}{12} =64$$

$$\Rightarrow x=8\sqrt 3$$

We know that the circumradius of a regular hexagon = side of regular hexagon

Hence r=8$$\sqrt 3$$$$\approx$$ 13.856

Therefore r=14 ( nearest integer)

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