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

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 ?

- $12$
- $14$
- $16$

Geometry

Triangle

Hexagon

But try the problem first...

Answer:$14$

Source

Suggested Reading

PRMO (2018) Problem 7

Pre College Mathematics

First hint

Show \(\triangle PAB \)is similar to \(\triangle PFC \).

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

Second Hint

The circumradius of a regular hexagon = side of regular hexagon

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

Final Step

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)

- https://www.cheenta.com/area-of-square-and-circle-amc-8-2011-problem-25/
- https://www.youtube.com/watch?v=BXWGKh01nOw

Contents

[hide]

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

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 ?

- $12$
- $14$
- $16$

Geometry

Triangle

Hexagon

But try the problem first...

Answer:$14$

Source

Suggested Reading

PRMO (2018) Problem 7

Pre College Mathematics

First hint

Show \(\triangle PAB \)is similar to \(\triangle PFC \).

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

Second Hint

The circumradius of a regular hexagon = side of regular hexagon

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

Final Step

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)

- https://www.cheenta.com/area-of-square-and-circle-amc-8-2011-problem-25/
- https://www.youtube.com/watch?v=BXWGKh01nOw

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