Math IOQM

Q1) Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides. Q2) Prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from all its vertices.

  Q1) Determine real x,y s.t. x^8 + y^8 = 8xy - 6 S1) LHS increases much faster than RHS, so only for some small values both sides will equal. Also when we have number of equations less than variables, here are few possible approaches: 1. make perfect squares 2. Some inequalities 3. Some useful substitutions. This problem does look like AM GM case since we have power of 8 and 8 terms: x^8,y^8,1....1(6 times). Also AM GM works only for non-negative numbers, which is true here. x^8 + y^8 + 1 ... 1)/8 >= (x^8.y^8.1 ... 1)^(1/8) = |xy| => x^8 + y^8 + 1 ... 1 >= 8|xy| >= 8xy But our question says they are equal which is only possible when all individual numbers are equal. So x^8 = y^8 = 1 => x,y = -+1 and upon checking they do satisfy. Answer. Q2. S2. x^2.y.z = 4^2 y^2.zx = 9^2 z^2.xy = 16^2 => x^4.y^4.z^4 = 4^2.9^2.16^2 = 2^4.3^4.4^4 => xyz = 2.3.4 = 24 => x = 16/24 = 2/3 y = 81/24 = 27/8 z = 256/24 = 32/3

 Q) Does there exist a sequence of integers a1,a2.... such that for each integer d != 0 there are exactly 2025 distinct pairs of indices i,j for which ai - aj = d? Answer: Yes But how to construct such a sequence? Before that let me mention that there is nothing special about 2025 in this problem. It could have been any other number. Let's try a way: If we target d = 1 Then we can simply have a sequence: 1,2....2026 and we have exactly 2025 pairs of integers with diff = 1. Now for d = 2: We already have 2024 such pairs in 2,3...2026, we just need to add one more. Let's add the number 2028 so that 2026,2028 is the required new pair. Now for d=3: We already have 2023 such pairs in 2,3....2026, and 2025,2028 is another. So total 2024. Just add 2031 so that 2028,2031 is another pair and we are done. Now for d=4 we have 2022 pairs in the intial block. then 2024,2028. then add 2035 to pair with 2031. And then 2039 for 2035. Now we have met target for d=4. So basically at each step se...

Also known as: The Midline Property of Angle Bisectors Here is the statement: When you drop perpendiculars from a vertex to the internal and external bisectors of the other two angles, the four feet always lie on a single straight line (the midline of the triangle). Proof: Internal angle bisector of B is the line which sits midway between AB and BC. AB reflected in this will overlap with BC. If you take A's reflection in this which maps to A', A' will lie somewhere on BC. And the internal bisector of B will be the perpendicular bisector of AA'. Let Q be the midpoint of AA'. Similarly A'' is A's reflection in angle B's external bisector. A'' will lie on BC. We can again show that if AA'' has P as its midpoint then BP is the perpendicular bisector AA''. Let B be (0,0). A' = (x1,0), A'' = (x2,0), A = (x1,y1) y co-ordinates of P and Q will be same  = y1/2 Midpoints of AB and AC will also have the same y co-ordinate. So...

If D,E,F are the feet of altitudes from A,B,C to BC,CA,AB then sides of DEF are anti parallel to ABC. What does it mean? Let's focus on BCEF quadrilateral which is cyclic. Angle AEF = B and angle AFE = C. If these angles were swapped then EF would be || to BC. But since they are in the exact opposite order it's called anti parallel. Same can be shown for other sides. Also in the triangle DEF, extending the argument above: Angle D = 180 - 2A Angle E = 180 - 2B Angle F = 180 - 2C Assuming that ABC is an acute angle triangle.

Q1. If the medians BE and CF are equal in a △ABC prove that AB=AC.  S1. Let G be the centroid and let CG = 2x = BG. Let EG = FG = x. Now triangles EGC and FGB are congruent by SAS. Why? Since EG = FG, CG = BG and angle FGB = angle CGE because they are vertical angles. => FB = EC And BF = FA since F is midpoint of AB, similarly AE = EC. => AF + FB = AE + EC => AB = AC H.P. Q2. Let (D, E, F) be the feet of the altitude from (A, B, C) in a (\triangle ABC). Prove that the perpendicular bisector of (EF) also bisects (BC). S2. BCEF is a cyclic quadrilateral with BC as diameter. Perpendicular bisector of EF will pass through the center of the circle since EF is a chord. Center lies on the midpoint of BC as it is a diameter. Hence perpendicular bisector of EF will pass through the midpoint of BC. H.P. Q3. If  S S  is the circumcentre of a  △ A B C △ A BC  and  D , E , F D , E , F  are the feet of the altitudes of  △ A B C △ A BC  th...