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 ?
Geometry
Triangle
Hexagon
But try the problem first...
Answer:$14$
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)
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 ?
Geometry
Triangle
Hexagon
But try the problem first...
Answer:$14$
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)