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.

24. Let P be an interior point of a triangle ABC whose sidelengths are 26, 65, 78. The line through P parallel to BC meets AB in K and AC in L. The line through P parallel to CA meets BC in M and BA in N. The line through P parallel to AB meets CA in S and CB in T. If KL, MN, ST are of equal lengths, find this common length. Solution:  Draw the triangle and see that PKBT, PMCL,PSAN are parallelograms. Let: PT = KB = x PM = LC = y PK = BT = z KL = MN = ST = l (the common length) Triangle PTM ~ ABC y/65 = (26-l)/26 Similarly, Triangle NKP ~ ABC (l-y)/65 = (78-l)/78 Solving this you get l = 30. But l should be less than each triangle side so it should be less than 26. So no solution.

 What is the number of triples (a, b, c) of positive integers such that (i) a<b<c< 10 and (ii) a, b, c, 10 form the sides of a quadrilateral? Solution: 4 sides of a rectangle have the property that sum of any 3 sides is larger than the 4th side. Similar to what we have in a triangle. And it can be generalized to any polygon. So: a + b + c > 10 a + b + 10 > c Which will anyway hold since c <= 9 a + c + 10 > b Which will anyway hold since b <= 8 c + b + 10 > a Which will anyway hold since a <= 7 So only the first one is worth examining. Now a,b,c have to be integers between 1 to 9. So we have 9C3 ways of choosing a,b,c. Out of which we have to remove those triplets which violate the first condition. 9C3 = 9.8.7/3.2 = 84 Now list down triplets which violate  a + b + c > 10 i.e. check for a,b,c for which a + b + c <= 10 Those are 11 triplets:  1,2,3 1,2,4 1,2,5 1,2,6 1,2,7 1,3,4 1,3,5 1,3,6 1,4,5 2,3,4 2,3,5 So answer 84 - 11 = 73.

17. Suppose the altitudes of a triangle are 10, 12 and 15. What is its semi-perimeter?  Solution: First, denote the area of the triangle by Δ, and let its sides be \(a,b,c\) opposite the altitudes \(h_a=10,\;h_b=12,\;h_c=15\). Then by definition of altitude, \[ a=\frac{2Δ}{h_a},\quad b=\frac{2Δ}{h_b},\quad c=\frac{2Δ}{h_c}. \] Hence the semiperimeter is \[ s=\frac{a+b+c}{2} =\frac1{2}\Bigl(\frac{2Δ}{10}+\frac{2Δ}{12}+\frac{2Δ}{15}\Bigr) =Δ\Bigl(\tfrac1{10}+\tfrac1{12}+\tfrac1{15}\Bigr) =Δ\cdot\frac{1}{4} =\frac{Δ}{4}. \] On the other hand, by Heron’s formula, \[ Δ^2 =s(s-a)(s-b)(s-c), \] and one checks that \[ s-a=\frac{Δ}{4}-\frac{2Δ}{10}=\frac{Δ}{20},\quad s-b=\frac{Δ}{12},\quad s-c=\frac{7Δ}{60}. \] Plugging in gives \[ Δ^2 =\frac{Δ}{4}\;\frac{Δ}{20}\;\frac{Δ}{12}\;\frac{7Δ}{60} =\frac{7\,Δ^4}{57600} \;\;\Longrightarrow\;\; Δ^2=\frac{57600}{7} \;\;\Longrightarrow\;\; Δ=\frac{240}{\sqrt7}. \] Therefore \[ s=\frac{Δ}{4}=\frac{240}{4\sqrt7} =\frac{60}{\sqrt7} =\frac{60\sqrt7}{7} \...

In a rectangle ABCD, E is the midpoint of AB; F is a point on AC such that BF is perpendicular to AC; and FE perpendicular to BD. Suppose BC = 8sqrt(3). Find AB.   Solution: 1. Set up a simple grid. Put the rectangle so that \[ A=(0,0),\quad B=(x,0),\quad C=(x,y),\quad D=(0,y), \] where \(x=AB\) (unknown) and \(y=BC=8\sqrt3\). 2. Locate the midpoint \(E\). Since \(E\) is the midpoint of \(AB\), \[ E =\Bigl(\tfrac x2,\,0\Bigr). \] 3. Find the foot \(F\) of the perpendicular from \(B\) onto \(AC\). - The diagonal \(AC\) has slope \[ m_{AC} \;=\;\frac{y-0}{x-0}\;=\;\frac{y}{x}\,, \] so any line through \(B\) perpendicular to \(AC\) must have slope \[ m_{BF} \;=\;-\frac1{m_{AC}}\;=\;-\frac{x}{y}\,. \] - Thus the line \(BF\) has equation \[ y \;=\;-\,\frac{x}{y}\,\bigl(X - x\bigr) \quad\bigl(\text{since it passes through }(x,0)\bigr). \] - Meanwhile \(AC\) itself i...

10. What is the greatest possible perimeter of a right-angled triangle with integer side lengths if one of the sides has length 12? Solution: If we have to maximize the perimeter, we should not keep 12 as the hypotenuse. That way hypotenuse will be longer than 12 and the perimeter will be more as compared to keeping 12 as hypotenuse. Let hypotenuse be y and the other leg be x. x^2 + 12^2 = y^2 y^2 - x^2 = 144 (y + x)(y - x) = 144 Since x,y are both integers the problem now reduces to finding integer factors of 144. Let's try them one by one. y + x = 144 y - x = 1 => 2y = 145 => y  = 72.5, x = 71.5, not integer move on. y + x = 72 y - x = 2 => 2y = 74 => y = 37, x = 35 valid solution y + x = 48 y - x = 3 => 2y = 51 => y = 25.5, x = 22.5, not integer move on. Should we try more factors now? Not really. Even if x,y solve for integer values, their values are constantly decreasing as the factors are becoming smaller. So y = 37, x = 35 is the right answer. Perimeter = 3...

9. A \(2 \times 3\) rectangle and a \(3 \times 4\) rectangle are contained within a square without overlapping at any interior point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square? Solution: There are 2 optimal ways to arrange the rectangles.  First do it vertically like this so that '2' and '4' are aligned. 2x3 4x3 So it becomes 6x3 _ _ _  |       | |       | |       | |       | |       | |       | The smallest square containing this has to be 6x6 with area 36. Second, '2' and '3' are aligned: 2x3 3x4 Now it becomes roughly 5x4 and the surrounding square has to be 5x5 = 25. Answer: 25

15. Let \( XOY \) be a triangle with \( \angle XOY = 90^\circ \). Let \( M \) and \( N \) be the midpoints of legs \( OX \) and \( OY \), respectively. Suppose that \( XN = 19 \) and \( YM = 22 \). What is \( XY \)? Solution: Let \( a = OX^2 \) and \( b = OY^2 \). Since \( \angle XOY = 90^\circ \), by the Pythagorean Theorem: \[ XY^2 = OX^2 + OY^2 = a + b \] Since \( N \) is the midpoint of \( OY \), \[ XN^2 = OX^2 + \left(\frac{OY}{2}\right)^2 = a + \frac{b}{4} \] \[ XN = 19 \Rightarrow XN^2 = 361 \Rightarrow a + \frac{b}{4} = 361 \quad \text{(1)} \] Since \( M \) is the midpoint of \( OX \), \[ YM^2 = OY^2 + \left(\frac{OX}{2}\right)^2 = b + \frac{a}{4} \] \[ YM = 22 \Rightarrow YM^2 = 484 \Rightarrow b + \frac{a}{4} = 484 \quad \text{(2)} \] Multiply equation (1) by 4: \[ 4a + b = 1444 \quad \text{(3)} \] Multiply equation (2) by 4: \[ a + 4b = 1936 \quad \text{(4)} \] Solve equations (3) and (4) simultaneously: Multiply equation (3) by 4: \[ 16a + 4b = 5...

10. In a triangle ABC, X and Y are points on the segments AB and AC, respectively, such that AX : XB = 1 : 2 and AY : YC = 2 : 1. If the area of triangle AXY is 10 then what is the area of triangle ABC? [PRE-RMO–2014] Solution (using a construction trick): Let's try to construct a simple triangle satisfying these conditions. Let's put right angle at A. Now area of AXY = zk = 10 Area of ABC = 9zk/2 = 45 Solution(using area formula): One of the area formula of triangle is Area = 1/2(side1 length)(side 2 length)(sin(included angle)) Here the included angle is A which is common in both. So Ratio of area = ratio of sides multiplied. So area ABC = 3/2 * 3/1 * 10 = 45

4. In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 17. What is the greatest possible perimeter of the triangle? Solution: Sides are a,3a,17 a + 3a > 17 => 4a > 17 => a > 17/4 17 + 3a > a => 17 > -2a 17 + a > 3a => 17 > 2a => a < 17/2 So 17/4 < a < 17/2 => 4.25 < a < 8.5 Since a is integer max value of a is 8 Hence the perimeter = 8 + 24 + 17 = 49

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 \[ ...