Math IOQM

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...

21. Find the number of ordered triples (a, b, c) of positive integers such that abc = 108. Prerequisite: Stars and bars theorem . Solution : To find the number of ordered triples ( a , b , c ) (a, b, c) of positive integers such that a b c = 108 abc = 108 , we can follow these steps: Step 1: Prime Factorization of 108 First, we need to factor 108 into its prime factors. 108 = 2 2 × 3 3 108 = 2^2 \times 3^3 Step 2: Assigning Prime Factors to a a , b b , and c c We can express a a , b b , and c c in terms of their prime factors: a = 2 r × 3 q , b = 2 t × 3 u , c = 2 z × 3 v a = 2^r \times 3^q,\quad b = 2^t \times 3^u,\quad c = 2^z \times 3^v where r , t , z r, t, z are the powers of 2 and q , u , v q, u, v are the powers of 3. Step 3: Setting Up the Equations From the product a b c = 108 abc = 108 , we can equate the powers of the prime factors: r + t + z = 2 (for the power of 2) r + t + z = 2\quad \text{(for the power of 2)} q + u + v = 3 (for th...

“If 1,2,3 are roots of  x 4 + a x 2 + b x = c x^4 + a x^2 + b x = c , find c c ." Answer: 36 Solution: simply put 1,2,3 so you get 3 equations in 3 vars and solve for c. A simpler solution: By Vieta's formulas Sum of roots = 0, so 4th root = -6 Product of roots = -c = -36 => c = 36

15. Integers 1, 2, 3, ..., n where n > 2 are written on a board. Two numbers m , k (such that 1 < m < n , 1 < k < n ) are removed and the average of the remaining numbers is found to be 17. What is the maximum sum of the two removed numbers? [PRE-RMO – 2017] Solution: Let two numbers moved be 1, 2. Then average of remaining (n − 2) numbers x ˉ 1 = n ( n + 1 ) 2 − 3 ÷ ( n − 2 ) = n 2 + n − 6 2 ( n − 2 ) = ( n + 3 ) ( n − 2 ) 2 ( n − 2 ) = n + 3 2 \bar{x}_1 = \frac{n(n+1)}{2} - 3 \div (n - 2) = \frac{n^2 + n - 6}{2(n - 2)} = \frac{(n + 3)(n - 2)}{2(n - 2)} = \frac{n + 3}{2} If two numbers be n , n − 1 n, n - 1 , then x ˉ 2 = n ( n + 1 ) 2 − ( 2 n − 1 ) ÷ ( n − 2 ) = n 2 − 3 n + 2 2 ( n − 2 ) = n − 1 2 \bar{x}_2 = \frac{n(n + 1)}{2} - (2n - 1) \div (n - 2) = \frac{n^2 - 3n + 2}{2(n - 2)} = \frac{n - 1}{2} Now, the removed numbers are m , k m, k , then new average is 17 i.e., 17 = n ( n + 1 ) 2 − ( m + k ) ÷ ( n − 2 ) 17 = \frac{n(n + 1)}{2} - (m + k)...

16. Five distinct 2-digit numbers are in a geometric progression. Find the middle term. Any integer‐valued 5‐term G.P. of two‐digit numbers forces the common ratio to be either 3 2 \tfrac32 or 2 3 \tfrac23 . Indeed, one finds: Ratio 3 2 \tfrac32 . Starting with a a , the terms are a , 3 2 a , 9 4 a , 27 8 a , 81 16 a . a,\quad \tfrac32 a,\quad \tfrac94 a,\quad \tfrac{27}{8}a,\quad \tfrac{81}{16}a. All must be integers and two‐digit. In particular, 81 16 a \tfrac{81}{16}a must be an integer, so a a must be a multiple of 16 16 . The only two‐digit multiple of 16 16 that keeps all terms under 100 100 is a = 16 a=16 . That yields 16 ,    24 ,    36 ,    54 ,    81 , 16,\;24,\;36,\;54,\;81, whose middle term is 36. 36. Ratio 2 3 \tfrac23 . Now the terms are a , 2 3 a , 4 9 a , 8 27 a , 16 81 a , a,\quad \tfrac23 a,\quad \tfrac49 a,\quad \tfrac{8}{27}a,\quad \tfrac{16}{81}a, and similarly a a must be a multiple of 81 81 . Taking a = 81 a=81 stays two‐digit an...

5. Let u , v , w u, v, w be real numbers in geometric progression such that u > v > w u > v > w . Suppose u 40 = v n = w 60 u^{40} = v^n = w^{60} . Find the value of n n . If u , v , w u,v,w form a geometric progression with common ratio r , r, then we can write v = u r , w = u r 2 . v = u r, \quad w = u r^2. The conditions u 40 = v n = w 60 u^{40} = v^n = w^{60} become u 40 = ( u r ) n and u 40 = ( u r 2 ) 60 . u^{40} = (ur)^n \quad\text{and}\quad u^{40} = (ur^2)^{60}. From the second equation, ( u r 2 ) 60    =    u 60   r 120    =    u 40 , (ur^2)^{60} \;=\; u^{60}\,r^{120} \;=\; u^{40}, which simplifies (dividing both sides by u 40 u^{40} ) to u 20   r 120    =    1       ⟹       u 20 = r − 120       ⟹       u = r − 6 . u^{20}\,r^{120} \;=\; 1 \;\;\Longrightarrow\;\; u^{20} = r^{-120} \;\;\Longrightarrow\;\; u = r^{-6}. Plug this into the first equation, ( u r ) n    =    ( r − 6 ⋅ r ) n    =    r − 6 n + n    =    r − 5 n (ur)^n \;=\; (r^{-6}\cdo...

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...