SOLUTIONS EGMO AND UCD ENRICHMENT PROGRAMME IN MATHEMATICS SELECTION TEST FEBRUARY 08. In triangle ABC, P is a point on AB, Q is a point on AC and X is the point of intersection of the line segments PC and QB. The quadrilateral AP XQ has area. The triangles QXC and P X B have area 5 and respectively. What is the area of the triangle ABC? Solution: Let x, y and z be the areas of the triangles AQX, AP X and B XC respectively. Thus x + y = area of AP XQ =. Now area(aqb) area(aqx ) = area(qbc ) area(qxc ) = 5 x = z + 5 = z = 5 5 x 5. Similarly On comparing, we obtain area(apc ) area(ap X ) = area(pbc ) area(p X B) = 9 y = z + = z = 9 x. z = 5 x 5 = 9 x = 0 = 5( x) 9x x( x) = x 50x + 00. Thus x = 5/ or 0 and since x, x = 5/. So y = 3/ and z = area(b XC ) = 5. Thus area(abc ) = area(ap XQ) + area(b X P) + area(c XQ) + area(b XC ) = + + 5 + 5 = 5.. Are there any positive integers n and m such that the integer 3 n + 35 m is a prime number? Prove your assertion. Solution: No: All such numbers are composite since 3 = 5 and 35 = 5 5 and hence 3 n + 35 m = 5n + 5 5m = ( n ) 5 + (5 m ) 5 = ( n + 5 m )( n 3n 5 m + n 5 m n 5 3m + 5 m ). Because m,n, the first factor is larger than. To see that the second factor is larger than as well, we consider two cases. In the first case, we assume 5 m n. We then obtain n 3n 5 m + n 5 m n 5 3m + 5 m = n + ( 5 3m + n 5 m)( 5 m n) n 8. The second case to consider is 5 m < n. In this case we have n 3n 5 m + n 5 m n 5 3m + 5 m = ( 3n + n 5 m)( n 5 m) + 5 m > 5 m 65. 3. How many different pairs of integers (x, y) satisfy the equation Write down 3 such pairs. Solution: There are 8 pairs of integer solutions: The left-hand side factors and the equation can be written 0x + 9x y + y = 5? (x + 3y)(5x + 7y) = 5. Since x, y are integers so are a := x + 3y and b := 5x + 7y. And ab = 5. Thus there are 8 possible values of (a, b): (a,b) = (,5),(5,),(, 5),( 5, ),(3,5),(5,3),( 3, 5),( 5, 3). Now given any pair (a,b) of integers, I claim that there is a unique pair x, y satisfying x + 3y = a and 5x + 7y = b: The two equations give: 0x +5y = 5a and 0x +y = b. Subtracting gives y = 5a b. Substituting this into either equation and solving for x gives x = 3b 7a. So x, y are uniquely determined by a,b proving the claim. (Note that these formulae also guarantee that x and y are integers whenever a and b are.) So there are precisely 8 different pairs (x, y) of solutions. For example, the pair (a, b) = (, 5) gives the solution x = 38, y = 5. The pair (a,b) = (3,5) gives the solution x = 6, y = 5. The pair (a,b) = (5,3) gives the solution x = 6, y = 9.
. Suppose n is a positive integer, such that all the digits of 7n, written in decimal notation, are 0 s and s. Find the smallest such n. Solution. As 7n = 8 9n = 9 8n, we deduce 7n must be a multiple of 8 and of 9. As 7n is a multiple of 8, then the number composed of the last three digits must be a multiple of 8. This follows because 000 is already a multiple of 8, so any higher digits will have no impact in the residual modulo 8. We are told all the digits of 7n are 0 s and s. There are 8 combinations for the last 3 digits: 000, 00, 00, 0, 00, 0, 0 and. Of these, only 000 is a multiple of 8. Therefore, the last 3 digits of 7n must be 000. Now as 7n is a multiple of 9, all its digits must add up to a multiple of 9 (because all powers of 0 are congruent to mod 9). The digits are all 0 s and s, so there must be at least 9 s. The smallest number satisfying these constraints is 7n =,,,000. Thus, by long division, the smallest n is n =,53,09,875. Alternatively, we may first observe that n must be divisible by 5 3, leading to three zeros at the end of 7n. Indeed, 7 is divisible by, thus 7n is even, so it ends in zero. Hence n must be divisible by 5. Then divide 7 by and n by 5 and apply the same reasoning two more times. We write n = 5 3 m and need to find the smallest m for which 9m consists of 0 s and s only. In the same way as above we see that we need 9 digits at least, so the smallest possibility for 9m is = 0000. The only divisions to be carried out are now /3 = 37 and 0000/3 = 333667. For full mars it is acceptable to say that n = 5 3 37 333667 without multiplying it out. 5. Emma writes the numbers,,3,...,9 into the cells of a 3 3 table (placing a different number in each cell). Then, she performs a series of moves as follows. In each move, she chooses an arbitrary square of the table, and either increases by or else decreases by all four numbers of that square. After performing a sequence of moves, Emma notices that all 9 numbers in the table are equal to some number n. Find, with proof, all possible values of n. Solution. n = 5. Let A, B, C, D denote the corner cells of the table. A B C D First, note that the sequence of moves performed is equivalent to performing the following four moves only: () adding some integer a to all four cells of the square containing the cell A; () adding some integer b to all four cells of the square containing the cell B; (3) adding some integer c to all four cells of the square containing the cell C ; () adding some integer d to all four cells of the square containing the cell D. Also, denote by x the initial number in the central cell of the table. Then, the final number in the central cell must be n = x + a + b + c + d. Since the initial sum of all the numbers in the table is + + + 9 = 5, one can easily see that the final sum of the numbers is 5 + (a + b + c + d). Thus we have 9n = 5 + (a + b + c + d) = 5 + (n x), or 5n = 5 x. It follows that x is divisible by 5 and since x 9 we get x = 5 and therefore n = 5 is the only possibility. It remains to show that n = 5 indeed is possible. One example in which Emma can obtain all entries equal to 5 is: 3 5 7 6 9 8 where she will choose a = 3,b =,c = and d = 3.
6. The non-zero real numbers a,b,c,d satisfy the equalities a + b + c + d = 0, a + b + c + d + abcd = 0. Find, with proof, all possible values of the product (ab cd)(c + d). Solution. From we deduce that So, Using the fact that a + b = (c + d) yields a + b + c + d + abcd = 0 bcd + cda + dab + abc =. = bcd + cda + dab + abc = cd(b + a) + ab(c + d). = (c + d)(ab cd) and so (ab cd)(c + d) = for all admissible values of a, b, c, d. 7. Let S be a set of 08 points in the plane, no three of which are collinear. Let P be a point not contained in any line segment that connects two points from S. Prove that the number of triangles that contain P, with vertices at three different points in S, is even. Solution. Consider moving the point P in the plane, while eeping the points of S fixed. If P lies outside the convex hull of S, it is contained in zero triangles (an even number). Notice that if we move P in a region without intersecting any of the line segments joining pairs of points in S, the number of triangles it is contained within cannot change. So we consider what happens as we move between regions (by crossing such a line segment). We can get to any region defined by the lines of S by crossing one line segment at a time. Consider line segment AB, where A,B S. Its extension divides the plane into two half-planes. Assume that one half-plane R contains points of S and the other half-plane T contains l points (so that + l = 06). As P crosses from R into T, it will no longer be contained in triangles of the form ABC where C R, but it will now be contained in all triangles of the form ABC where C T. Therefore, the total number of triangles with vertices in S which contain P will change by exactly l, which is an even number since and l have the same parity (from + l = 06). Since this total number of triangles is initially zero and can only change by an even number, it must be even, and so the result is proved. 8. (a) Prove that for any positive real numbers x, y we have x 3 + y 3 x y + x y. (b) Prove that for any real numbers 0 x, y, z we have 3 + x 3 + y 3 + z 3 x + y + z + x + y + z. Solution. (a) We have x 3 + y 3 (x y + x y ) = (x + y)(x y) 0. (b) Using the above inequality we have Similarly we have + x 3 + y 3 + x y + x y ( + x y) + ( + x y ) (x + y) + (y + x). Adding the above three inequalities we obtain the conclusion. + y 3 + z 3 z + y + y + z + z 3 + x 3 z + x + x + z. Alternatively, we may start with the observation that for 0 x we have ( x )( x) 0. This immediately implies that + x 3 x + x. The same is true for y and z. Adding these three inequalities gives the desired result of (b). Note. Both parts can be solved by use of the rearrangement inequality. For part (a) use sequences {x, y} and {x, y }, and for part (b) use sequences {,,, x, y, z} and {,,, x, y, z }. The only constraint we impose is x, y, z 0. 3
9. Let A,B,C,D be four distinct points on a circle (in this order). Let P be the intersection of AD and BC, and let Q be the intersection of AB and CD. Prove that the angle bisectors of DPC and AQD are perpendicular. Solution. Denote the intersection of the bisectors by I. Let E,F be the intersection points of PI with AB and CD and denote by G, H the intersection point of Q I with BC and AD respectively. Also, Hence, HPI = DPC DC = AB. PH I = HDQ + HQD = ADC + AQD HPI + PH I = DC AB This implies that H I P = 90 o so PI QI. + AB + BC + AD BC = AB + BC + AD BC. = AB + BC + AD + D A = 90 o. An alternative solution goes as follows. We use the notation α = EPB = EPA,β = BQG = CQG and γ = EBP = GBQ. Because the exterior angles at triangles EBP and BGQ are equal to the sum of the opposite interior angles and the angle sum in quadrilateral E IGB is 360 we get 360 = PIQ + QE I + PGI + EBG = PIQ + α + γ + β + γ + (80 γ), hence 80 = PIQ + α + β + γ. On the other hand, from the cyclic quadrilateral ABCD we obtain 80 = D AB + DCB = α + γ + β + γ considering exterior angles at triangles ABP and BCQ. This implies α + β + γ = 90, hence the desired PIQ = 90. 0. Let {S n : n = 0,,...} be a sequence defined by S 0 = and S n = S n + n for n. Show that S n n, for all n. Solution. We prove this by induction on n. The statement clearly holds for n =, where equality holds since S = S 0 + = =. Let us suppose that the inequality holds at n = for some, so that: S. Observe that: ( ) = + >
As, we have 0, so we can tae the positive square root of each side of the inequality: > Doubling and rearranging, this implies that: < Adding this to the inductive hypothesis, we have: S = S + + =. Thus the hypothesis also holds for n =. As we already demonstrated the result numerically for n =, it therefore holds for all positive integers n. The inductive step could, alternatively, be carried out as follows. We start with the observation that ( ) < ( ) for all integers. If > 0, both sides are positive and we can tae the positive square roots to get ( ) <, i.e. ( ) + <. Dividing by, produces + <. Using the definition of S and the inductive hypothesis now gives S = S + + <. 5