Math IOQM

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

8. The equations x 2 − 4 x + k = 0 x^2 - 4x + k = 0 and x 2 + k x − 4 = 0 x^2 + kx - 4 = 0 , where k k is a real number, have exactly one common root. What is the value of k k ? [PRE-RMO – 2015] Solution: Let the common root be 'b'. First equation has roots a,b and the second one has b,c. a + b = 4 ab = k a = 4-b = k/b => 4b - b^2 = k => b^2 = 4b - k ____[1] Similarly, b + c = -k bc = -4 c = -k-b = -4/b => b^2 + kb = 4 => b^2 = 4 - kb________[2] From [1] and [2] 4 - kb = 4b - k => b = 1 => k = 3 Answer: 3 Solution 2: Suppose the two quadratics have a common root α \alpha . Then { α 2 − 4 α + k = 0 , α 2 + k α − 4 = 0. \begin{cases} \alpha^2 - 4\alpha + k = 0, \\ \alpha^2 + k\alpha - 4 = 0. \end{cases} Subtracting one equation from the other gives ( α 2 + k α − 4 )    −    ( α 2 − 4 α + k )    =    ( k + 4 )   α    −    ( k + 4 )    =    ( k + 4 ) ( α − 1 ) = 0. (\alpha^2 + k\alpha - 4) \;-\; (\alpha^2 - 4\alpha + k) \;=\; (k + 4)\,\alpha \;-\; (k+4...