Math IOQM

a,b,c,d,e are real numbers. If a+1 = b+2 = c+3 = d+4 = e+5 = a+b+c+d+e+3 then a^2 + b^2 + c^2 + d^2 + e^2 = ? Sol: b = a-1, c = a-2, d=a-3, e = a-4 a+1 = 5a-7 a = 2 b = 1,c=0,d=-1, e=-2 a^2 + b^2 + c^2 + d^2 + e^2 = 10 Ans: 10

11. Three real numbers x , y , z x, y, z are such that x 2 + 6 y = − 17 x^2 + 6y = -17 , y 2 + 4 z = 1 y^2 + 4z = 1 , and z 2 + 2 x = 2 z^2 + 2x = 2 . What is the value of x 2 + y 2 + z 2 x^2 + y^2 + z^2 ? Solution: Add all LHS and RHS and make whole squares of x,y,z You will get: (x + 1)^2 + (y+3)^2 + (z+2)^2 = 0 So each term is 0. That gives you the exact values of x,y,z.

1. Diagonals of a kite meet at 90 Deg. 2. The diagonal joining vertices between unequal sides gets bisected by the other diagonal. Proof: In a kite ABDC, AB = AC DB = DC Diagonals AD and BC meet at E. ABD and ACD are congruent by SSS. So Angle  BAD = Angle CAD Which is same as saying that angle BAE = angle CAE Now ABE and ACE are congruent by SAS. So angle AEB = AEC Since AEB + AEC = 180 so AEB = AEC = 90, H.P. Using this congruence of ABE and ACE, we also prove that BE = EC, i.e. the diagonal joining vertices between  unequal sides gets bisected by the other diagonal.

Carol was given three numbers and was asked to add the largest of the three to the product of the other two. Instead, she multiplied the largest with the sum of the other two, but still got the right answer. What is the sum of the three numbers? Solution: If the numbers are a,b,c then ab + c = ac + bc Doing trial and error, this seems to have multiple solutions, for e.g. a = b = c = 0 a = c = 1, b = 0 a = b = c = 1 etc. So I am adding another constraint which was not given in the original question: These numbers are consecutive natural numbers. Then you can do a,a+1,a+2. Now you will get a quadratic equation in a. Which will give a = 0,-3 And then we choose a = 0 and next 2 numbers are 1,2.

Let Akbar and Birbal have n marbles, where n>0. Akbar says to Birbal, "If I give you some marbles you will have twice as many as I will have." Birbal says to Akbar, "If I give you some marbles you will have thrice as many as I will have." What is the minimum possible value of n for which the above statements are true? Solution: Suppose Akbar gives Birbal some marbles leaving Akbar with k marble and Birbal with 2k marbles. This implies that n=k+2k for some integer k i.e. n is divisible by 3. Similarly, after Birbal gives Akbar some marbles leaving Birbal with m marbles and Akbar with 3m marbles, we have that n=m+3m=4m for some integer m, i.e. n is divisible by 4. So minimum value of n = LCM(3,4) = 12. Source

Prerequisites: Divisibility by 11 . Solution 1: 11p + 13n = 1000 Now we can do: 13n = 1000 - 11p n = (1000 - 11p)/13 For n to be an integer, 1000 - 11p should be divisible by 13. So we can do p=1,2,3... until we find the first number divisible by 13. But wait, isn't it easier to check divisibility by 11? Right? So let's do the other way round. p  = (1000 - 13n)/11 Now put n = 1, Numerator is 987 which is not divisible by 11 since sum of odd digits - sum of even digits is not divisible by 11. Keep doing it and you will get the first success at n = 5 when numerator becomes 935. So Rs. 935 on pens and Rs. 65 on notebooks. So Rs. 65 is the minimum I can spend on notebooks. From here, if I want to spend Rs. X more on notebooks then I will have to spend Rs. X on pens. So X has to be a multiple of 11,13. In fact, it has to be the LCM. Which is 143. So keep subtracting 143 from 935 as long as it doesn't become negative. 143*6 = 858 So you can subtract 6 times. Hence total solutions...

You can check out a simpler solution here. This one is a bit more complicated and uses modulo arithmetic. The problem states: A pen costs ₹11. A notebook costs ₹13. We need to find the number of ways to spend exactly ₹1000 on these items. Step 1: Formulating the equation Let: x x be the number of pens. y y be the number of notebooks. Then the equation to satisfy is: 11 x + 13 y = 1000 11x + 13y = 1000 where x , y x, y are non-negative integers. Step 2: Finding integer solutions We solve for x x in terms of y y : x = 1000 − 13 y 11 x = \frac{1000 - 13y}{11} For x x to be an integer, ( 1000 − 13 y ) (1000 - 13y) must be divisible by 11. Let's determine valid values for y y by checking divisibility: 1000 − 13 y ≡ 0 ( m o d 11 ) 1000 - 13y \equiv 0 \pmod{11} Since 1000 ≡ 10 ( m o d 11 ) 1000 \equiv 10 \pmod{11} and 13 y ≡ 2 y ( m o d 11 ) 13y \equiv 2y \pmod{11} , we get: 10 − 2 y ≡ 0 ( m o d 11 ) 10 - 2y \equiv 0 \pmod{11} 2 y ≡ 10 ( m o d 11 ) 2y \equiv...

Answer: 91 One by one keep X,Y,Z in middle and see what are the possible options. Only X---3--Z--7--Y and Z--3--X--10--Y are possible. So answer = 7*13 = 91

\[ x^2 + 4y^2 - 2xy - 2x - 4y - 8 = 0 \] \[ \Rightarrow x^2 - 2x(y+1) + 4y^2 - 4y - 8 = 0 \] \[ \Rightarrow x^2 - 2x(y+1) + y^2 + 2y + 1 + 3y^2 - 6y - 9 = 0 \] \[ \Rightarrow (x - (y+1))^2 + 3(y-1)^2 = 12 \] \[ \Rightarrow (x - y - 1)^2 + 3(y - 1)^2 = 12 \] Let \( X = x - y - 1 \) and \( Y = y - 1 \) \[ \Rightarrow X^2 + 3Y^2 = 12 \] Since \( x, y \in \mathbb{Z} \), it follows that \( X, Y \in \mathbb{Z} \). \[ (X^2, Y^2) \in \{(0,4), (9,1), (3,4)\} \] \[ (X, Y) \in \{(0,2), (0,-2), (3,1), (3,-1), (-3,1), (-3,-1)\} \] Evaluate \( (x, y) \) for each \( (X, Y) \): \[ X = x - y - 1, \quad Y = y - 1 \] \[ (x, y) \in \{(4,3), (0,-1), (6,2), (1,0), (0,2), (-2,0)\} \] **6 possible integer pairs.**

a = b-c ---1 b = c-d ---2 c = d-a ---3 Add LHS and RHS a+b+c = b-a =>a+c = -a => c = -2a ---4 From 1,4 => b = -a similarly, c = -2a d = -a => a/b + b/c + c/d + d/a = 1/2 Answer: 1/2

Suppose that \( 4^{x_{1}} \) = 5, \( 5^{x_{2}} \) = 6 \( \ldots \)and \( 127^{x_{124}} \) = 128. What is the value of the product X1.X2....X124 ? Answer: 7/2.

  Multiplying R×O+M=46 term to term by M we have M×R×O+M^2=46M but M×R×O=240 so M^2−46M2+240=0 will give us M=6 and M=40 as a whole numbers In the same way R^2+R×M×O=64R so R^2−64R+240=0 will give us R=4 and R=60 To obtain the O values, substituting into M×R×O=240 we obtain M=6,R=4,O=10 M=6,R=60,O= M=40,R=4,O= M=40,R=50,O= Anser: R+M+O = 20

Source I think positive integers start at 1, don’t they? So the minimum value would be 1… y=1 then x = 100–3*1 = 97 We can continue this: y=2 x = 100 - 3*2 = 94 … until y = 33 x = 100 - 3*33 = 1 for higher values of y there’s no positive x: y = 34 x = 100 - 3*34 = 100 - 102 = -2 So, there are 33 pairs, from y=1 to y =33. If you said “non negative integers” instead of positive integers, there would be 34 pairs: from y=0 to y=33. It’s just that easy. Why did I choose the y to try values? Well, the x can be solved as an integer for every y : x = 100 - 3*y 100 - 3*y is always an integer. But if you try some x and solve for y you can’t find always an integer: 3y = 100 - x y = (100 - x) / 3 and (100 - x) / 3 is not always an integer.

A country has a food deficit of 10 % . Its population grows continuously at a rate of 3 % per year. Its annual food production every year is 4% more than that of the last year. Assuming that the average food requirement per person remains constant, prove that the country will become self-sufficient in food after n years, where n is the smallest integer bigger than or equal to (ln 10 - ln 9)/ (ln ( 1.04 ) - 0.03). Solution: Key point to note here is that population is growing continuously whereas food production is growing annually. So P_n = P_0.e^(rt) F_n = F_0.(1 + r/100)^t

Question: Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen? Solution 1: We start by noting that each of the 4 boxes contains 3 red and 2 blue distinct balls. In any valid selection of 10 balls, from each box at least one red and one blue must be chosen. Step 1. Determine How Many Balls Are Chosen from Each Box Let x i x_i be the number of balls chosen from box i i (for i = 1 , 2 , 3 , 4 i=1,2,3,4 ). We have: x 1 + x 2 + x 3 + x 4 = 10 x_1+x_2+x_3+x_4 = 10 Since from each box at least one red and one blue must be chosen, we must choose at least 2 balls from each box: x i ≥ 2 for  i = 1 , 2 , 3 , 4. x_i \ge 2 \quad \text{for } i=1,2,3,4. Define y i = x i − 2 y_i = x_i - 2 (so y i ≥ 0 y_i \ge 0 ). Then: y 1 + y 2 + y 3 + y 4 = 10 − 8 = 2. y_1+y_2+y_3+y_4 = 10 - 8 = 2. The n...

Answer: 1

Question 2, Pre RMO 2018-19(IOQM): In a quadrilateral ABCD, it is given that AB = AD = 13, BC= CD = 20, BD = 24. If r is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to r?. Prerequisites: 1. Incenter, incircle and angle bisectors in triangle/polygons 2. Kite diagonals Solution : ABCD is a kite. So, A C ⊥ B D AC \perp BD . So, B O = O D = 12  cm BO = OD = 12 \text{ cm} . A O = 13 2 − 12 2 = 5  cm AO = \sqrt{13^2 - 12^2} = 5 \text{ cm} O C = 20 2 − 12 2 = 16  cm OC = \sqrt{20^2 - 12^2} = 16 \text{ cm} ar ( □ A B C D ) = 1 2 × A C × B D \text{ar}(\square ABCD) = \tfrac{1}{2} \times AC \times BD = 1 2 × 21 × 24 = \tfrac{1}{2} \times 21 \times 24 = 252  cm 2 = 252 \text{ cm}^2 r = Δ S = 252 33 = 7.63  cm r = \frac{\Delta}{S} = \frac{252}{33} = 7.63 \text{ cm} Closest integer = 8 = 8 .

  Pre RMO 2018-19(IOQM) question 1: 1. A book is published in three volumes, the pages being numbered from 1 onwards. The page numbers are continued from the first volume to the second volume to the third. The number of pages in the second volume is 50 more than that in the first volume, and the number of pages in the third volume is one and a half times that in the second. The sum of the page numbers on the first pages of the three volumes is 1709. If  n  is the last page number, what is the largest prime factor of  n ? Prerequisites: 1. Divisibility by 11 .

IOQM 2023 Q5: In a triangle ABC , let E be the midpoint of AC and F be the midpoint of AB . The medians BE and CF intersect at G . Let Y and Z be the midpoints of BE and CF respectively. If the area of triangle ABC is 480 , find the area of triangle GYZ . Answer 10:  Solution: 1. 3 medians intersecting at centroid divide the triangle into 6 triangles of equal area. 2. GBC has 2 of those triangles so its area is 1/3 of the total area. 3. Look at GYZ. GB is 2/3 of the median BE. BY is 1/2 of the median BE. So GY = 2/3 - 1/2 = 1/6BE. GYZ and GBC are similar with side ratio of 1:4. So [GYZ] = 1/16[GBC] = 1/16 * 1/3 [ABC] = 480/16*3 = 10. Prerequisites: Centroid and median properties in triangle . Similarity criterion for triangles.

 AAA(which is same as AAS or ASA) SAS SSS RHL Note that these are same as triangle congruence criteria but for congruence we ensure that sides are equal and for similarity we ensure that corresponding sides have the same ratio.

 1. What is the criteria for divisibility by 11? Sum of odd digits - sum of even digits = 0 or 11k

Centroid: 1. 3 Medians divide triangle in 6 equal parts.  Proof . 2. 3 Medians are concurrent and centroid divides median in 2:1. Proof . 3. How do 2 medians divide triangle? Relation between areas? Use 1. to come up with this relation. 4 areas in 2:1. 4. Centroid is always inside the triangle.

  Incenter: 1. Why are 3 angle bisectors concurrent? Proof: A. Angle bisector by definition is equidistant from the 2 adjacent sides.  B. So the point where 2 angle bisectors meet is equidistant from all 3 sides. C. Hence the 3rd angle bisector will also pass through that point. 2. Since a triangle will always have an incenter, it will always have an incircle. 3. Easy to prove that the radius of incenter r = A/s where A is area of triangle and s = perimeter/2. Steps: A. Since incenter is equidistant from all sides, let's drop perpendiculars to sides, its length is r. B. All sides are tangents to the circle and hence meet those 3 perpendiculars. C. Now 3 sub triangles, area of each is a*r/2 where a is the corresponding side length. D. Add all these sub areas and equate to main area to get the proof. 4. Same formula applies to any polygon which has an incircle. 5. Not all polygons have an incenter though. They may or may not have it. 6. But all regular polygons have an incenter....