Math IOQM

15. Let \(A_1,B_1,C_1,D_1\) be the midpoints of the sides of a convex quadrilateral \(ABCD\), and let \(A_2,B_2,C_2,D_2\) be the midpoints of the sides of the quadrilateral \(A_1B_1C_1D_1\). If \(A_2B_2C_2D_2\) is a rectangle with side lengths 4 and 6, what is the product of the lengths of the diagonals of \(ABCD\)? Answer: 208 Solution: Assume that ABCD is a rectangle and draw all the points. Construct the diagram and use TMT(Triangle Midsegment Theorem) and symmetry. You will realized that diagonals of A1B1C1D1 are 8,12 and they are also the sides of ABCD. So diagonal length of ABCD will be Sqrt(208). Hence the answer 208.

 Paper Let \(ABC\) be an equilateral triangle of side length \(s\). Let \(P\) and \(S\) be points on \(AB\) and \(AC\), respectively, and let \(Q\) and \(R\) be points on \(BC\) such that \(PQRS\) is a rectangle. If \[ PQ = \sqrt3\,PS \quad\text{and}\quad \text{Area}(PQRS)=28\sqrt3, \] what is the length of \(PC\)? Solution: First find that PS = \(2\sqrt7\) and PQ = \(2\sqrt21\). Now APS and ABC are similar so APS is also equilateral. Hence AP = PS = AS = \(2\sqrt7\). Let's say SC = x, then RC = x*cos(60 deg) = x/2 and SR = x * sqrt(3)/2 So SC = 4*sqrt(7) And RC = 2*sqrt(7) So each side of the triangle is 6*sqrt(7). QC = QR + RC = 2*sqrt(7) + 2*sqrt(7) Now PC = Sqrt(PQ^2 + QC^2) = 14

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 \[ ...