Math IOQM

  Below is exactly the chain-of-thought I would go through if this dropped onto my scratch paper in a contest or an exam. I’m writing it slowly so you can see the cues that tell me *what* to complete the square in and *why* those particular shifts appear. --- 1. Look for a **single-variable** square you can force The moment you see a quadratic in two variables, $$ x^{2}+4y^{2}-2xy-2x-4y-8=0, $$ pick one variable (almost always the one with coefficient 1 in its squared term, because that makes the algebra cleaner). Here $x^{2}$ already has coefficient 1, so treat **$y$ as a constant** for a moment and complete the square in **$x$** exactly the way you would in one-variable algebra. Write the $x$-part together: $$ x^{2} \;-\; 2xy \;-\; 2x. $$ It’s a quadratic in $x$ whose linear coefficient is $-2(y+1)$. So $$ x^{2}-2(y+1)x \;=\; \bigl(x-(y+1)\bigr)^{2} \;-\;(y+1)^{2}. $$ > **Why $y+1$?** > Because the standard completing-the-square move says: $x^{2} - 2px = (...

20. PS is a line segment of length 4 and O is the midpoint of PS. A semicircular arc is drawn with PS as diameter. Let X be the midpoint of this arc. Q and R are points on the arc PXS such that QR is parallel to PS and the semicircular arc drawn with QR as diameter is tangent to PS. How can I get the area of the region QXROQ bounded by the two semicircular arcs? Answer: 2*PI - 2 Solution: Image taken from here . Now, the big circle has area 4*Pi. Since RQ is the diameter of the smaller circle, it subtends 90 degree on its circumference. So Angle QOR is right angle. So the line segments QO and OR will capture 1/4th the area of the completed blue circle which is Pi.--------[1] Now we just need to find the area between green  lines and the red circular arcs below them. If you see the area captured by line segments YQ and YO in the red circle, it will be 1/4th of the completed red circle. Again the same reason that angle QYQ is right angle. Radius of the red circle is sqrt(2). So the c...

18. What is the sum of the squares of the roots of the equation \(x^2 - 7\lfloor x \rfloor + 5 = 0\)? Here \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to x. For example, \(\lfloor 3.4 \rfloor = 3\) and \(\lfloor -2.3 \rfloor = -3.\) Solution: x^2 + 5 = 7[x] LHS > 0 so RHS has to be positive. So x has to be >= 1. LHS is quadratic and RHS is linear. So after a certain value of x, LHS will overtake RHS and RHS will not be able to catch up. If that value is not too big, we can enumerate all the possible cases. If we put x = 7, we see that LHS > RHS and for all subsequent values of x, RHS won't be able to catch up. So we will try only upto x = 7. Also, note that RHS is integer. So LHS has to be integer. So x^2 has to be integer since 5 is already integer. So x has to be square root of an integer. Now we start enumeration. First rewrite the equation as: \(x^2 = 7[x] - 5\). 1 x^2 = 7 - 5 = 2 => x = sqrt(2). We don't consider t...

17. Let \(x_1, x_2, x_3\) be the roots of the equation \(x^3 + 3x + 5 = 0\). What is the value of the expression $$ \left( x_1 + \frac{1}{x_1} \right)\left( x_2 + \frac{1}{x_2} \right)\left( x_3 + \frac{1}{x_3} \right)? $$ Solution: $$ x^2 + 3x + 5 = 0 $$ By Vitae's formulae: $$ x_1 + x_2 + x_3 = 0 $$ $$ 3 = x_1 x_2 + x_2 x_3 + x_3 x_1 $$ $$ -5 = x_1 x_2 x_3 $$ So the given sum becomes: $$ \frac{{(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)}}{(x_1)(x_2)(x_3)} $$ We know the denominator here is -5. So let's work on numerator. First expand the numerator, then use these 2 substitutions: $$ (x_1 + x_2 + x_3)^2 = x_1^2 + x_2^2 + x_3^2 + 2(x_1x_2 + x_2x_3 + x_3x_1) $$ and similarly $$ (x_1x_2 + x_2x_3 + x_3x_1)^2 = ... $$ to get the final answer -29/5.

Q16. Let N be the set of natural numbers. Suppose f : N -> N is a function satisfying the following conditions: (a) f(mn) = f(m)f(n) (b) f(m) < f(n) if m < n (c) f(2) = 2 What is the value of $$\sum_{k=1}^{20} f(k)?$$ Answer: 210 Solution: Since f(1) < f(2), f(1) = 1 (the only natural number < 2). f(4) = f(2).f(2) = 4 f(2) < f(3) < f(4) so f(3) = 3 is the only choice. f(6) = f(3).f(2) = 6 so f(5) = 5. It's clear that f(n) = n is the function definition here. So we just to have to compute the sum of first 20 natural numbers.

14. \( O \) and \( I \) are the circumcentre and incentre of \( \triangle ABC \) respectively. Suppose \( O \) lies in the interior of \( \triangle ABC \) and \( I \) lies on the circle passing through \( B, O, \) and \( C \). What is the magnitude of \( \angle BAC \) in degrees? You can apply a trick here. If you can construct a triangle which satisfies the conditions given in the question, you can answer using that. Here if you construct an equilateral triangle, O will lie in the interior of the triangle and I will lie on the circle passing through O,B,C since in the equilateral triangle O and I are same. So the answer will simply be 60 degrees. Detailed solution copied from here : We know, because \( I \) is the incenter, \[ \angle IBD = \frac{B}{2}, \quad \angle ICD = \frac{C}{2} \] \[ \Rightarrow \angle BIC = \pi - \left( \frac{B}{2} + \frac{C}{2} \right) = \pi - \left( \frac{\pi - A}{2} \right) = \frac{\pi}{2} + \frac{A}{2} \] But, \( \angle BIC = \angle BOC \), because c...

$$ \text{If } \frac{1}{\sqrt{2011} + \sqrt{2011^2 - 1}} = \sqrt{m} - \sqrt{n}, \text{ where } m \text{ and } n \text{ are positive integers, what is the value of } m + n? $$ Solution: Multiply up and down with $$ {\sqrt{2011} - \sqrt{2011^2 - 1}} $$ to get: $$ {\sqrt{2011} - \sqrt{2011^2 - 1}} $$ If this is equal to: $$ \sqrt{m} - \sqrt{n} $$ then squaring both sides we get: $$ 2011 - \sqrt{2011^2 - 1} = m + n - 2\sqrt{mn} $$ so $$ 2011 - \sqrt{2012.2010} = m + n - 2\sqrt{mn} $$ and $$ 2011 - 2\sqrt{1006.1005} = m + n - 2\sqrt{mn} $$ and $$ 1006 + 1005 - 2\sqrt{1006.1005} = m + n - 2\sqrt{mn} $$ so m = 1006, n = 1005, m + n = 2011.

10. ABCD is a square and AB = 1. Equilateral triangles AYB and CXD are drawn such that X and Y are inside the square. What is the length of XY? Solution: In the triangle AYB, perpendicular from Y will cut AB at its midpoint E. Similarly, in the triangle CXD, perpendicular from X will cut CD at its midpoint F. Length of those perpendiculars will be EY = XF = Sqrt(3)/2. Now EY + XF > 1because Y lies inside CXD and X lies inside AYB. So EY + YX + XF = 1 and we need find XY. But it's not clear how. So let's try another approach. If we add EY and XF then the sum would be 1 + XY. Why? Because when we add EY and XF, the segment XY is added twice. So EY + XF = sqrt(3) = 1 + XY => XY = sqrt(3) - 1 => Answer Image source  

  In rectangle \(ABCD\), \(AB = 5\) and \(BC = 3\). Points \(F\) and \(G\) are on line segment \(CD\) so that \(DF = 1\) and \(GC = 2\). Lines \(AF\) and \(BG\) intersect at \(E\). What is the area of \(\triangle AEB\)? Solution: Triangles EAB and EFG are similar by AAA. Let's say height of EFG is h. Then height of EAB is 3 + h. So AB/FG = (3+h)/h => 5/2 = (3+h)/h => 5h = 6 + 2h => h = 2 => 3 + h = 5 So area of EAB = (5*5)/2 = 12.5

7. In △ A B C \triangle ABC , we have A C = B C = 7 AC = BC = 7 and A B = 2 AB = 2 . Suppose that D D is a point on line A B AB such that B B lies between A A and D D and C D = 8 CD = 8 . What is the length of BD? [PRE-RMO–2012] Solution: Since ABC is an isosceles triangle. Let's say E is the midpoint of AB. Then CAB will be a right angle. Let BD = x. CE = 7^2 - 1^2 = 8^2 - (1+x)^2 => 48 = 64 - (1+x)^2 => x + 1 = 4 => x = 3 Ans. 3

Q6: A postman has to deliver five letters to five different houses. Mischievously, he posts one letter through each door without looking to see if it is the correct address. In how many different ways could he do this so that exactly two of the five houses receive the correct letters? Solution: Let's do it via an example. Let's say 5 houses are A,B,C,D,E. Let's pick 2 correct houses A,B and compute how many ways are there to deliver remaining 3 letters incorrectly. Whatever the answer, we will multiply it with 5C2(=10) since there are 5C2 ways to choose the 2 correct houses. So now distribute 3 letters incorrectly to C,D,E. To the first house C there are 2 ways to give the wrong letter(D or E). Let's pick D. Now to remaining 2 houses D,E there is only 1 way to give the incorrect letter. So in total there are 2x1x1 ways. D,E,C E,C,D So answer is 10x2 = 20.

 2. A triangle with perimeter 7 has integer side lengths. What is the maximum possible area of such a triangle? Possible side lengths: 1,1,5 => invalid (since sum of 2 sides should be more than the 3rd) 1,2,4 => invalid 1,3,3 => valid 2,2,3 => valid So only 2 combinations are valid => 1,3,3 and 2,2,3. Compute area by heron's formula for both the combinations. The larger area is 3.sqrt(7)/4 and that's the answer.

5. Let S n = n 2 + 20 n + 12 S_n = n^2 + 20n + 12 , n n a positive integer. What is the sum of all possible values of n n for which S n S_n is a perfect square? [PRE-PRMO–2012] Solution Answer: 3 + 13 = 16 Let n 2 + 20 n + 12 = u 2 n^2 + 20n + 12 = u^2 ( n + 10 ) 2 − 88 = u 2 (n + 10)^2 - 88 = u^2 ( n + 10 − u ) ( n + 10 + u ) = 88 (n + 10 - u)(n + 10 + u) = 88 As 88 = 88 × 1 = 44 × 2 = 22 × 4 = 11 × 8 88 = 88 \times 1 = 44 \times 2 = 22 \times 4 = 11 \times 8 Using each of the pair of factors of 88, we can get different values of n n and u u . Of all the values, we get n = 3 n = 3 and n = 13 n = 13 as positive integer solutions. Thus, sum of all possible values of n n is 3 + 13 = 16 3 + 13 = 16

Let x 1 , x 2 , x 3 x_1, x_2, x_3 be the roots of the equation x 3 + 3 x + 5 = 0. x^3 + 3x + 5 = 0. What is the value of the expression ( x 1 − 1 x 1 ) ( x 2 − 1 x 2 ) ( x 3 − 1 x 3 ) \left( x_1 - \frac{1}{x_1} \right) \left( x_2 - \frac{1}{x_2} \right) \left( x_3 - \frac{1}{x_3} \right) [PRE-RMO – 2012] Prerequisites: Vieta's formulas And (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) Source Sol: x 3 + 0 x 2 + 3 x + 5 = 0 x^3 + 0x^2 + 3x + 5 = 0 have roots x 1 , x 2 , x 3 x_1, x_2, x_3 ⇒ x 1 + x 2 + x 3 = 0 ; x 1 x 2 + x 2 x 3 + x 3 x 1 = 3 ; x 1 x 2 x 3 = − 5 \Rightarrow x_1 + x_2 + x_3 = 0 ; \quad x_1x_2 + x_2x_3 + x_3x_1 = 3 ; \quad x_1x_2x_3 = -5 ( x 1 + 1 x 1 ) ( x 2 + 1 x 2 ) ( x 3 + 1 x 3 ) ⇒ ( x 1 x 2 + x 1 x 2 + x 2 x 1 + 1 x 1 x 2 ) ( x 3 + 1 x 3 ) (x_1 + \frac{1}{x_1}) (x_2 + \frac{1}{x_2}) (x_3 + \frac{1}{x_3}) \Rightarrow (x_1x_2 + \frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{1}{x_1x_2})(x_3 + \frac{1}{x_3}) = ( x 1 x 2 x 3 + x 1 x 2 x 3 + x 3 x 1 x 2...

\[ x^2 + 4y^2 - 2xy - 2x - 4y - 8 = 0 \] \[ \Rightarrow x^2 - 2x(y+1) + 4y^2 - 4y - 8 = 0 \] \[ \Rightarrow x^2 - 2x(y+1) + y^2 + 2y + 1 + 3y^2 - 6y - 9 = 0 \] \[ \Rightarrow (x - (y+1))^2 + 3(y-1)^2 = 12 \] \[ \Rightarrow (x - y - 1)^2 + 3(y - 1)^2 = 12 \] 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.