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.

 1. if parallel side lengths of a trapezium are a,b find the formula for its area given the height between a,b is 'h'. 2. Now if you put a = 0 what do you get? what does it mean?

 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 \[ ...

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

Let abc be a three digit number with nonzero digits such that a^2 + b^2 = c^2. What is the largest possible prime factor of abc? Solution: Only single digit Pythagorean triplet is 3,4,5 so the number could be 345 or 435. 345 = 3 x 5 x 23 435 = 3 x 5 x 29 Answer: 29

7. On a clock, there are two instants between 12 noon and 1 PM, when the hour hand and the minute hand are at right angles. The difference in minutes between these two instants is written as a + b/c , where a, b, c are positive integers, with b < c and b/c in the reduced form. What is the value of a + b + c ? Solution: Assume that clock is made up of 360 degrees. Hour hand moves 360/12 = 30 degrees per hour => 30/60 degrees per minute => 0.5 degrees per minute. Minute hand moves 360 degrees per hour => 360/60 degrees per minute => 6 degrees per minute. Speed of minute hand relative to hour hand  = 6 - 0.5 = 5.5 degrees per minute. The angle between them would become 90 after 90/5.5 minutes. And it would become 270 after 270/5.5 minutes. Difference between those 2 instants  = (270-90)/5.5 = 180/5.5 = 32 + 8/11 a = 32, b = 8, c = 11 a + b + c = 51

14. Find the smallest positive integer n ≥ 10 n \ge 10 such that n + 6 n + 6 is a prime and 9 n + 7 9n + 7 is a perfect square.  Solution: p = n + 6 9n + 7 = k^2 Solving we get k^2 = 9p - 47 Given n >= 10 So starting value of p is 17(16 is not prime). Each time we increase p by 1, k^2 increases by 9. p = 17, k^2 = 106 (not valid) keep trying for p = 23,29,31,37,41,43,47,53,59 at p = 59, k^2 = 484 => k = 22 n = 53 answer.

Q. One day I went for a walk in the morning at x x minutes past 5 O’ clock, where x x is a two-digit number. When I returned, it was y y minutes past 6 O’ clock, and I noticed that (i) I walked exactly for x x minutes and (ii) y y was a 2-digit number obtained by reversing the digits of x x . How many minutes did I walk? [PRE-RMO–2019] This is quite simple actually. Let x = ab and y = ba => x = 10a + b, y = 10b + a Now x minutes past 5 = 300 + x since 5 hours have 300 minutes. So 360 + y - (300 + x) = x => 60 + y = 2x => 30 + y/2 = x => 30 + (10b+a)/2 = 10a + b => 30 + 5b + a/2 = 10a + b => a = 2/19(30 + 4b) => a = (60 + 8b)/19 Now a,b are between 0,9 So just keep putting values of 'b' to find an answer. On b = 2, a = 4 So x = 42 and y = 24

Preparation: 1.  If x + y = 130 and their gcd is 10 find all possible pairs of (x,y). Given that x,y are positive integers(natural numbers). 2. If xy = 7 + x + y and x,y are positive integers find the pairs (x,y) satisfying this equation. 3. Complete a rectangle method. How do you factorize axy + bx + cy = d? Solution: Multiply both sides by a => a^2xy + abx + acy = ad => ax(ay + b) + acy = ad => ax(ay + b) + ay(c) = ad => ax(ay + b) + (ay + b)(c) - bc = ad => (ax + c)(ay + b) = ad + bc Now coming to the main question: 11. For natural numbers  x  and  y , let  ( x , y )  denote the greatest common divisor of  x  and  y . How many pairs of natural numbers  x  and  y  with  x ≤ y  satisfy this equation: xy = x + y + (x,y) Solution: Let gcd(x,y) = d xy = x + y + d xy - x - y = d (x-1)(y-1) = d + 1 Let a,b co primes such that, x = ad, y = bd Since x <= y, it means that a <= b. Also a,b are natur...

 Below is a short, purely “algebraic” argument (no explicit modular arithmetic) for the common test for divisibility by 7 : Take the last digit of the number, double it, and subtract that product from the remaining truncated part. If the result is divisible by 7, then the original number is also divisible by 7. In a formula: Let N N have last digit b b and truncated part a a . So N = 10 a + b N = 10a + b . The rule says: “ N  is divisible by 7”    ⟺    “ a − 2 b  is divisible by 7.” \text{“\(N\) is divisible by 7”} \;\Longleftrightarrow\; \text{“\(a - 2b\) is divisible by 7.”} We prove this equivalence purely via algebraic manipulation and the notion of “multiple of 7.” Forward Direction (If N N is divisible by 7, then a − 2 b a - 2b is also divisible by 7.) Assume 10 a + b 10a + b is a multiple of 7. That is, there exists some integer k k such that 10 a + b    =    7 k . 10a + b \;=\; 7k. Rewrite “ 10 a 10a ...

Divisibility “rule” for 13 (subtract 9 times the last digit) A known test for divisibility by 13 13 works by taking the last digit, multiplying it by 9 9 , and then subtracting that product from the remaining truncated part of the number. If the result is divisible by 13 13 , then the original number is divisible by 13 13 . Here’s how it looks for 533 533 : Split off the last digit ( 3 3 ) and the rest of the number ( 53 53 ). Multiply the last digit by 9 9 : 3 × 9 = 27 3 \times 9 = 27 . Subtract that from the truncated part: 53 − 27 = 26 53 - 27 = 26 . Since 26 26 is divisible by 13 13 (i.e., 26 = 13 × 2 26 = 13 \times 2 ), 533 533 is also divisible by 13 13 . Either approach quickly confirms that 533 = 13 × 41 533 = 13 \times 41 . Proof: Below is a short, purely “algebraic” proof of the rule A decimal number N N is divisible by 13 if and only if “take its last digit b b , multiply it by 9, and subtract that from the truncated part a a .” The resulting number...

Prerequisites: Divisibility by 13. 11. Find the largest value of a b a^b such that the positive integers a , b > 1 a, b > 1 satisfy a b ⋅ b a + a b + b a = 5329 a^b \cdot b^a + a^b + b^a = 5329 Solution: Rewrite the given equation a b   b a    +    a b    +    b a    =    5329 a^b\,b^a \;+\; a^b \;+\; b^a \;=\; 5329 as ( a b + 1 ) ( b a + 1 )    =    5329 + 1    =    5330. (a^b+1)(b^a+1)\;=\;5329 + 1 \;=\;5330. Thus if we set x = a b x=a^b and y = b a , y=b^a, then ( x + 1 ) ( y + 1 ) = 5330. (x+1)(y+1)=5330. Next one factors 5330 5330 (which is 2 × 5 × 13 × 41 2 \times 5 \times 13 \times 41 ) and looks for factor‐pairs ( p , q ) (p,q) of 5330 5330 such that x = p − 1 x=p-1 and y = q − 1 y=q-1 can both be positive perfect powers a b a^b and b a b^a . Among the possible divisors, the only workable pairs turn out to be ( p , q )    =    ( 65 ,   82 ) and ( 82 ,   65 ) , (p,q) \;=\; (65,\,82)\quad\text{and}\quad(82,\,65), which give x = 64 ,    y = 81 or x = 8...

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...

Q24. If N is the number of triangles of different shapes (i.e., not similar) whose angles are all integers (in degrees), what is N/100? Prerequisite: Stars and bars method . Example: Distribute 3 apples in 3 kids. (3+3-1)_C_(3-1) = 5_C_2 = 10 (0,0,3) => 3 variations (1,1,1) => 1 variation (01,2) => 6 variations Total: 3+1+6 = 10 Solution: By stars and bars method if we have to distribute 'n' apples in 'k' kids where a kid can receive 0 apples also, then the number of ways are: (n+k-1)_C_(k-1). In case of triangles we have to distribute 180 degrees in 3 angles but none of the angles can be 0. So if the original equation is x + y + z = 180, we rewrite it as X + Y + Z = 177 where X = x-1, Y = y-1 and Z = z-1 Now X,Y,Z can be 0 and we can apply stars and bars method. 179_C_2 = 15931 But these distributions treat (0,0,177) and (0,177,0) as different. The original problem statement specifically asks for triangles for different shapes, so we have to treat them as same...

 Five persons are to be seated around a circular table they are wearing badges of consecutive numbers 1,2,3,4,5. In how many ways can they be seated such that no two consecutive numbered badge wearing people are seated next to each other ? Solution: Try to arrange it manually. Let's first put one of 2,3,4 because each of them has 2 adjacent numbers which we need to avoid whereas for 1 and 5 there is only 1 number to avoid. Let's start with 2.       2 Now adjacent numbers can be only 4,5             2      4           5 Now 3 can't go next to 4 so:             2      4           5         1    3 If you flip 4,5 initially             2      5           4         3    1 So effectively, ther...

In how many ways can a pair of parallel diagonals of a regular polygon of 10 sides be selected? Hint: First try for an octagon. Visual solution . Solution : If the decagon is P 1 ⋯ P 10 P_1 \cdots P_{10} then there are two types of sets of parallel diagonals. One is typified by P 3 P 10 ,   P 4 P 9 ,   P 5 P 8 P_3P_{10},\ P_4P_9,\ P_5P_8 (parallel to the side P 1 P 2 P_1P_2 ) and the other by P 1 P 3 ,   P 10 P 4 ,   P 9 P 5 P_1P_3,\ P_{10}P_4,\ P_9P_5 and P 8 P 6 P_8P_6 . There are five sets of each type. If you are counting unordered sets of parallel diagonals, then the answer is 5 ( 3 2 ) + 5 ( 4 2 ) = 45.

20. What is the number of ordered pairs ( A , B ) (A, B) where A A and B B are subsets of { 1 , 2 , … , 5 } \{1, 2, \ldots, 5\} such that neither A ⊆ B A \subseteq B nor B ⊆ A B \subseteq A ? [PRE-RMO–2014] A wrong approach: Let's partition the numbers into 2 sets. First set has 1 element, second has 4 elements. Total ways to do this = 5C1 * 2 = 10. 5C1 since there are 5C1 ways to choose the set with 1 element. *2 because these are ordered pairs. Similarly, let's partition the numbers such that first set has 2 elements, second set has 3 elements. Again, 5C2*2 = 20. So total 30 ways. Why is this wrong? Because we didn't count the cases when a number is not present in either set + when a number is present in both the sets. For e.g. A = {2}, B = {3,4} is a valid case. A = {2,3}, B = {1,2} is also valid. So what's the correct approach? Via the “four‐color” method. Label each element of { 1 , 2 , 3 , 4 , 5 } \{1,2,3,4,5\} according to whether it is in neither ...