Math IOQM

19. In a triangle \( ABC \) with \( \angle BCA = 90^\circ \), the perpendicular bisector of \( AB \) intersects segments \( AB \) and \( AC \) at \( X \) and \( Y \), respectively. If the ratio of the area of quadrilateral \( BXYC \) to the area of triangle \( ABC \) is \( 13 : 18 \) and \( BC = 12 \), then what is the length of \( AC \)? Prerequisites: 1. If two triangles are similar, ratio of their areas is same as ratios of their sides squared. Why? Because their heights share the same ratio. a1/a2 = h1/h2 So area ratio = a1h1/a2h2 = a1(a1h2/a2)/a2h2 = a1^2/a2^2 Solution : Firstly, we can state that, \[ \frac{\text{Area of } A Y X}{\text{Area of } A B C} = \frac{5}{18} \tag{1} \] This is because it is given that, \[ \frac{\text{Area of } B X Y C}{\text{Area of } A B C} = \frac{13}{18} \] and, \[ \text{Area of } A B C = \text{Area of } A Y X + \text{Area of } B X Y C. \] **Next we can state that,** \[ \frac{A B^2}{4 \times A C^2} = \frac{5}{18} \tag{2} \] This is because \( A B C \) ...

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 \[ ...

5. There are n − 1 n - 1 red balls, n n green balls, and n + 1 n + 1 blue balls in a bag. The number of ways of choosing two balls from the bag that have different colors is 299. What is the value of n n ? [PRE-RMO–2013] Partially correct approach: Choose 1 red and 1 non-red ball = (n-1)(n+n+1) = (n-1)(2n+1) Choose 1 green and 1 non-green ball = (n)(n-1+n+1) = (n)(2n) Choose 1 blue and 1 non-blue ball = (n+1)(n+n-1) = (n+1)(2n-1) Adding all we get 6n^2 - 2 = 299 But wait, LHS is even and RHS is odd. Something is wrong here. Oh, gotcha! We have done double counting. We have counted each combination twice. For e.g., in the first line we counted red + green and red + blue combinations. In the second line, green + red and green + blue. In the third line, blue + red and blue + green. We were supposed to count each one only once. So, divide LHS by 2. 3n^2 - 1 = 299 => n^2 = 100 => n = 10

Question: What is the the maximum possible value of k for which 2013 can be written as sum of k consecutive positive integers? Solution: Suppose that we start with some a ∈ N a \in \mathbb{N} . Then we want that a + ( a + 1 ) + … + ( a + ( k − 1 ) ) = 2013. a + (a+1) + \ldots + (a+(k-1)) = 2013. Rewriting it as the sum of arithmetic series we get ( 2 a + k − 1 ) k 2 = 2013 or ( 2 a − 1 + k ) k = 4026. \frac{(2a + k - 1)k}{2} = 2013 \quad \text{or} \quad (2a - 1 + k)k = 4026. Note that 2 a − 1 + k ≥ k 2a - 1 + k \geq k . So k, (2a - 1 + k) are 2 factors of 4026 such that 'k' is the smaller factor. We need to find all factor pairs of 4026 and take the biggest smaller factor. Now we find factor pairs of 4026. The factor pairs of 4026 are: - (1, 4026) - (2, 2013) - (3, 1342) - (6, 671) - (11, 366) - (22, 183) - (33, 122) - (61, 66) The biggest smaller factor is 61. Answer: 61. 2a - 1 + k = 66 = 2a + 60 => a = 3 Cross check that the A.P. starting from 3 and having 61 terms...

4. Let  f ( x ) = x 3 − 3 x + b f(x) = x^3 - 3x + b and g ( x ) = x 2 + b x − 3 g(x) = x^2 + bx - 3 , where b b is a real number. What is the sum of all possible value of b b for which the equation f ( x ) = 0 f(x) = 0 and g ( x ) = 0 g(x) = 0 have a common root? [PRE-RMO – 2013] Solution: Suppose  x = α x=\alpha is a common root of f ( x )    =    x 3 − 3 x + b and g ( x )    =    x 2 + b x − 3. f(x) \;=\; x^{3}-3x+b \quad\text{and}\quad g(x) \;=\; x^{2}+bx-3. Then α 3 − 3 α + b    =    0 , \alpha^{3}-3\alpha + b \;=\; 0, α 2 + b   α − 3    =    0. \alpha^{2}+b\,\alpha - 3 \;=\; 0. From the second equation, provided α ≠ 0 \alpha \neq 0 , solve for b b : b    =    3 − α 2 α . b \;=\;\frac{3 - \alpha^{2}}{\alpha}. Substitute this into the first equation: α 3 − 3 α    +    3 − α 2 α    =    0 , \alpha^{3} - 3\alpha \;+\; \frac{3 - \alpha^{2}}{\alpha} \;=\; 0, and multiply through by α \alpha (again using α ≠ 0 \alpha \neq 0 ): α 4    −    3 α 2    +    ( 3  ...

3. It is given that the equation  x 2 + a x + 20 = 0 x^2 + ax + 20 = 0 has integer roots. What is the sum of all possible values of a a ? [PRE-RMO – 2013] Solution 1: x = (a ± √(a² - 80)) / 2 a² - 80 should be a perfect square. For eg. a = 9 ⇒ √1 a = -9 also works And x should be integer. So numerator, i.e. a ± √(a² - 80) should be even. If a is odd, (a² - 80) is odd and hence √(a² - 80) is odd ⇒ numerator = -a ± odd = -odd ± odd = even Similarly if a is even, (a² - 80) is even and hence √(a² - 80) is even ⇒ numerator = -a ± even = -even ± even = even So numerator is always even if (a² - 80) is a perfect square. And for each valid value of (a² - 80), there are 2 valid values of 'a'(plus and minus). So no matter how many valid values of 'a' are there, they will all add up to 0. Answer: 0 Solution 2: A convenient way to proceed is to let the integer roots be  r 1 r_1 and r 2 r_2 . Then by Vieta’s formulas, r 1 + r 2 = − a and r 1   r 2 = 20. r_1 + r_2 = -a \quad\te...

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.

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

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