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practice problems pending

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1) ABCD is a parallelogram. Through C, a straight line RQ is drawn outside the figure and AP, BQ, DR are drawn ⟂ to RQ. Show that: DR + BQ = AP . S1) [RMO] Key idea: DR || BQ so they can form a trapezium. Since LHS is DR + BQ which also appears in Trapezium midline theorem we will try to use Trapezium midline theorem which says this: If AB || CD in trapezium ABCD, then line EF joining midpoints of AD and BC is || to AB,CD and also EF = (AB + CD)/2. Let O be intersection of diagonals AC and BD in the ||gram ABCD. O is their midpoint. Let OM be perpendicular to RQ. DR || BQ and DBQR is a trapezium. OM || DR || BQ and OM goes through midpoint of BD => M is the midpoint of RQ. OM = (BQ + DR)/2 We are almost there now. If we could show that OM = AP/2, we are done. OM || AP In Triangle APC, it's a strong hint to use Triangle midpoint theorem. O is midpoint of AC and OM || AP => M is midpoint of PC. => OM = AP/2 H.P. Another solution: Drop a perpendicular DO from D to AP. DRPO is...
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practice problems pending

1. Show that 10^(2n-1) + 1 is divisible by 11 for all natural number 'n'. Solution: 10 = -1 mod 11 10^(2n-1) = (-1)^(2n-1) mod 11 = -1 mod 11 => 10^(2n-1) + 1 = 0 mod 11 H.P. 2. Show that 11^(n+2) + 12^(2n+1) is divisible by 133 for all natural 'n'. Solution: 133 = 7*19 So LHS should be div by both 7 and 19 Try 7 first: 11 mod 7 = 4, 12 mod 7 = 5 LHS = 4^(n+2) + 5^(2n+1) 5 mod 7 = -2 => 5^2 mod 7 = 4 mod 7 LHS = 16.4^n + 5.4^n = 21.4^n mod 7 = 0 mod 7 Now do the same with 19. LHS = 121.11^n + 12.12^2n 12^2n = 144^n 144 = 11 mod 19 144^n = 11^n mod 19 121 = 7 mod 19 LHS = 7.11^n + 12.11^n = 19.11^n = 0 mod 19 H.P. 3. Find all integers a such that the quadratic expression (x+a)(x+1991) + 1 can be factored as (x+b)(x+c), where b and c are integers. Solution: Since -b is a root of RHS =>  (-b+a)(-b+1991) = -1 Case 1: (-b+a) = 1 (-b+1991) = -1 => b = 1992, a = 1993 Case 2: (-b+a) = -1 (-b+1991) = 1 => b = 1990, a = 1989 Answer: a = 1989, 1993.
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practice problems pending

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.
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practice problems pending

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  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
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practice problems pending

 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...
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The Collinearity of Feet of Perpendiculars to Bisectors - pending

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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...
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Sides of Orthic triangle DEF are anti parallel to ABC

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.
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