1. The positive integers x1, . . . , x7 satisfy the conditions
x6= 144, xn+3= xn+2(xn+1+ xn) n = 1, 2, 3, 4.
Compute x7.
Solution: Multiplying the given equation for n = 1, 2, 3 and can-celing terms we get:
144 = x3(x1+ x2)(x2+ x3)(x3+ x4). [1]
Also from the given equality we get the following inequalities:
x4 = x3(x2+ x1)≥ 2x3
x5 = x4(x3+ x2)≥ 2x23
144 = x6≥ x5(x4+ x3)≥ 2x23(3x3)⇒ 144 ≥ 6x33⇒ x3= 1, 2.
Case 1: x3= 1.
By [1] , 144 = (x1+ x2)(x2+ 1)(x1+ x2+ 1). The pairs of factors of 144 that are consecutive integers are (1,2) (2, 3), (3, 4), and (8, 9).
Since x1+ x2and x1+ x2+ 1 are factors of 144 that are consecutive integers and since x1+ x2≥ 2, we have 3 subcases:
Subcase 1a) x1+ x2= 2 implies 6(x2+ 1) = 144 and so x2= 23;
x1 = −21; however, this is not a valid solution since the xi are positive integers.
Subcase 1b) x1+ x2= 3 implies 12(x2+ 1) = 144; x2= 11; x1=
−8; again, this is invalid.
Subcase 1c) x1+ x2= 8 implies 72(x2+ 1) = 144; x2= 1; x1= 7. Testing this possible solution:x4 = 8, x5 = 16, x6 = 144, so (x1, x2, x3) = (7, 1, 1) works. The value of x7= 3456.
Case2: x3= 2
144 = 2(x1+ x2)(x2+ 2)(2x1+ 2x2+ 2) implies 36 = (x1+ x2)(x2+ 2)(x1+x2+1). The pairs of factors of 36 that are consecutive integers are (1, 2), (2, 3), (3, 4). We then analyze as above:
Subcase 2a) x1+ x2= 2 implies x2= 4; x1 =−2; invalid solu-tion.
Subcase 2b) x1+ x2 = 3 implies x2 = 1, x1 = 2. Testing this solution:
x4 = 6, x5= 18, x6= 144, so (x1, x2, x3) = (2, 1, 2) also works. The value of x7= 3456.
The value of x7is thus 3456.
2. Solve the following system of equations in real numbers x, y, z:
3(x2+ y2+ z2) = 1
x2y2+ y2z2+ z2x2 = xyz(x + y + z)3.
Solution: First note that none of x, y, z,or (x + y + z) can equal 0, for otherwise, by the second equation, x = y = z = 0, which does not satisfy the first equation. Also note that xyz(x + y + z) =
x2y2+y2z2+z2x2 (x+y+z)2 ≥ 0.
For three real numbers a, b, c, we have (a−b)2+(a−c)2+(b−c)2≥ 0, or equivalently a2+ b2+ c2≥ ab + ac + bc, with equality if and only if a = b = c. So
1 = 3(x2+ y2+ z2) ≥ (x + y + z)2=x2y2+ y2z2+ z2x2 xyx(x + y + z)
≥ xy2z + x2yz + xyz2 xyz(x + y + z) = 1.
Hence all the inequalities in the above expression must be equalitites, so we must have x = y = z. So from the first equation we find that the possible triples (x, y, z) are (13,13,13) and (−13 ,−13 ,−13 ). Both satisfy the given equations, so they are the solutions.
3. [Corrected] In a tetrahedron ABCD, the medians of the faces ABD, ACD, BCD from D make equal angles with the corresponding edges AB, AC, BC. Prove that each of these faces has area less than or equal to the sum of the areas of the other two faces.
Solution: Let α ≤ 90◦ be the common angle; first suppose α 6= 90◦. If a, b, c are the lengths of BC, CA, AB and ma, mb, mc are the lengths of the medians from BC, CA, AB, respectively, then
the area of triangle DAB is 12mcc sin α. Note that the absolute value of the dot product between the vectors (D− (A + B)/2) and A − B is mcc cos α = 2 cot α[DAB], but also equals |DA2− DB2|. Thus the three areas [ABD], [ACD], [BCD] are proportional to |DA2− DB2|, |DA2− DC2|, |DB2− DC2|, and the desired inequality is ev-ident in this case.
Now suppose α = 90◦. Let x = ∠ADB, y = ∠BDC, z = ∠ADC (0 < x, y, z < 180). Note that since x, y, z are the angles of a trihedral angle, we have x + y > z, x + y > y, y + z > x and x + y + z≤ 360.
Since the area of triangle ADB is (AD)(BD)(sin x)/2 (and similar relations hold for the areas of BDC and ADC) and AD = BD = CD (since α = 90), we need to prove that sin x + sin y > sin z (and the analogous inequalities).
Using trigonometric identities, we have sin x + sin y = 2 sin(x + y
2 ) cos(x− y
2 ), sin z = 2 sinz 2cosz
2, so we need to prove
sin(x + y
2 ) cos(x− y
2 ) > sinz 2cosz
2.
Since x + y + z ≤ 360 =⇒ x+y2 ≤ 180 − z2. Note that since 0 <
z
2 ≤ x+y2 ≤ 180 − z2, we have sin(x+y2 ) > sinz2. Also, since x−y2 <
z
2, we have cos(x−y2 ) > cosz2 since the cosine function is strictly decreasing on [0, 180] . Multiplying the two inequalities together (which is legal since the expressions involved are positive), we get sin(x+y2 ) cos(x−y2 ) > sinz2cosz2, as desired.
4. The sequence a1, a2, . . . is defined by
a1= 0, an= abn/2c+ (−1)n(n+1)/2 n > 1.
For every integer k≥ 0, find the number of n such that 2k≤ n < 2k+1 and an= 0.
Solution: Let Bndenote the base 2 representation of n.
First, we will prove by induction that an is the number of 00 or 11 strings in Bn minus the number of 01 or 10 strings in the Bn. For
the base case n=1, we have B1 = 1 so a1= 0− 0 = 0. Assume that for k = 1, 2, ...n− 1, ak is the number of 00 or 11 strings in Bk minus the number of 01 or 10 strings in the Bk. First consider the cases when n ≡ 0, 3 (mod 4). Then Bn ends in 00 or 11, so an equals one plus the number of 00 or 11 strings in all but the digit of Bn minus the number of 01 or 10 strings in the all but the last digit of Bn. This latter number is given by abn2c. Thus
an= abn2c+ 1 = abn2c+ (−1)n(n+1)/2,
as desired. Similarly, for n≡ 1, 2 (mod 4), we get (since Bn ends in 01 or 10 and we are subtracting these)
abn2c −1 = abn2c+ (−1)n(n+1)/2, completing the induction.
So for a given integer k, we need to find the number of integers n such that 2k ≤ n < 2k+1and the number of 00 and 11 strings equals the number of 01 and 10 strings. Note that Bn has k + 1 digits.
For each Bn, we construct a new sequence Cn of 0’s and 1’s as follows: starting at the leftmost digit of Bn and working our way to the penultimate digit of Bn, we add to our sequence Cn the absolute value of the difference between that digit and the digit to its right.
For example, B11=1011 and C11=110. Since a 00 or 11 string in Bn
yields a 0 in Cn and a 01 or 10 string in Bn yields a 1 in Cn, we need to find the number of integers n such that 2k ≤ n < 2k+1 and the number of zeroes in Cnequals the number of ones. Since Cnhas k digits (one less than Bn), an equal number of zeroes and ones is impossible if k is odd. If k is even, then we can select k2 of the k digits equal to one; the rest of the k2 places we can set equal to zero.
From this sequence Cn, we can reconstruct Bn; it is easy to see that each sequence Cn corresponds to a unique value of Bn and hence a unique value of n. There are kk
2 ways to select the placement of the ones. So the answer is 0 if k is odd, kk
2otherwise.
5. Given a convex pentagon ABCDE with DC = DE and ∠BCD =
∠DEA = π/2, let F be the point on segment AB such that AF/BF = AE/BC. Show that
∠F CE = ∠F DE and ∠F EC = ∠BDC.
Solution: Let P be the intersection of AE and BC, and note that C, D, E, P are concyclic. Let Q and R denote the intersections of DA and DB, respectively, with the circumcircle of CDEP . Let G = QC∩ RE. We have ∠GCE = ∠ADE and ∠GEC = ∠BDC.
Also, by Pascal’s Theorem for the hexagon P CQDRE, A, G, B are collinear. So all we must do is show AG/GB = AE/BC and then we conclude F = G. We do this via the law of sines:
where the last step follows from the fact that ADE and BDC are right triangles.
6. Consider n points (n≥ 2) on a unit circle. Show that at most n2/3 of the segments with endpoints among the n chosen points have length greater than√
2.
Solution: Construct a graph on the given vertices by connect-ing every pair of points whose distance is greater than√
2. We will show that no K4(the complete graph on 4 vertices) exists. Suppose a K4 exists; call its vertices (in cyclic order) ABCD.
Edges of length greater than √
2 subtend a minor arc of greater than π/2 radians. So AB, BC, CD, DA each subtend minor arcs of greater than π/2 radians, and together they subtend more than 2π radians. This is a contradiction, since there are only 2π radians in a circle. That the graph has at most n32 edges now follows from Turan’s theorem.