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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 ?

Circumscribed circle of the hexagon
  • $12$
  • $14$
  • $16$

Key Concepts


Geometry

Triangle

Hexagon

Check the Answer


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........

finding radius of Circumscribed circle of the hexagon

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 ?

Circumscribed circle of the hexagon
  • $12$
  • $14$
  • $16$

Key Concepts


Geometry

Triangle

Hexagon

Check the Answer


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........

finding radius of Circumscribed circle of the hexagon

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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