Problem
Let ABCD be a convex cyclic quadrilateral . Suppose P is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least .If {PA,PB,PC,PD} = {3,4,6,8}.What is the maximum possible area of ABCD?
Topic
Geometry
Contest
Pre Regional Math Olympiad
2019
Reading
Mathematical circle .
Sequential Hints
The first step is very clear, you have to some how locate the point P .
claim : The intersection of the diagonal is the required point .
Proof : suppose that there is a point P' other than P (the intersection of diagonal) .
now try to apply triangular inequality in \(\triangle\)PBD & \(\triangle\)APC.
(do the rest yourself , and complete the proof)
so , the area of [ABCD]=\(\frac{1}{2}\)\(\sin\theta\)(PA*PB+PB*PC+PC*PD+PD*PA)
now, the area will maximum when \(\sin\theta\)=1
so , just put the value of PA, PB, PC,PD
You get the answer 55
Watch video
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.
Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.
Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.