PRMO 2013 question 8

8. Let \(AD\) and \(BC\) be the parallel sides of a trapezium \(ABCD\). Let \(P\) and \(Q\) be the midpoints of the diagonals \(AC\) and \(BD\). If \(AD = 16\) and \(BC = 20\), what is the length of \(PQ\)?

Preparation:
A. Prove the TMT(Triangle Midsegment Theorem), i.e. if we join midpoints of two sides of a triangle, this new line will be parallel to the third line of the triangle and be half its length.
A1. There are 2 ways to prove it. One using Co-ordinate geometry and another using simple geometry.

B. Now come to a trapezium. Prove that if we join the midpoints of the diagonals, this new segment will be parallel to the 2 parallel sides of the trapezium. Look here for an example proof. You can also prove it using co-ordinate geometry.

Solution:
Now extend the proof in B. to calculate the length of various midsegments and find PQ.
Answer will be 2.

Alternate beautiful solution:

You can place the trapezoid so that \[ A=(0,0),\ D=(16,0),\quad B=(b,h),\ C=(b+20,h). \] Then \[ P=\text{mid}(A,C)=\Bigl(\tfrac{b+20}2,\tfrac h2\Bigr),\quad Q=\text{mid}(B,D)=\Bigl(\tfrac{b+16}2,\tfrac h2\Bigr), \] so \[ PQ=\Bigl|\tfrac{b+20}2-\tfrac{b+16}2\Bigr|=2. \] Hence \(\displaystyle PQ=2\).

Comments

Popular posts from this blog

Combinatorics DPP - RACE 6 - Q16 pending discussion

Geometry practice problems

Pre RMO 2018(IOQM), Question 2 incircle quadrilateral