1. Find all real numbers m such that the equation
(x2− 2mx − 4(m2+ 1))(x2− 4x − 2m(m2+ 1)) = 0 has exactly three different roots.
Solution: Answer: m = 3. Proof: By setting the two factors on the left side equal to 0 we obtain two polynomial equations, at least one of which must be true for some x in order for x to be a root of our original equation. These equations can be rewritten as (x− m)2= 5m2+ 4 and (x− 2)2= 2(m3+ m + 2). We have three ways that the original equation can have just three distinct roots:
either the first equation has a double root, the second equation has a double root, or there is one common root of the two equations.The first case is out, however, because this would imply 5m2+ 4 = 0 which is not possible for real m.
In the second case, we must have 2(m3+ m + 2) = 0; m3+ m + 2 factors as (m+1)(m2−m+2) and the second factor is always positive for real m. So we would have to have m =−1 for this to occur. Then the only root of our second equation is x = 2, and our first equation becomes (x + 1)2 = 9, i.e. x = 2,−4. But this means our original equation had only 2 and -4 as roots, contrary to intention.
In our third case let r be the common root, so x− r is a factor of both x2− 2mx − 4(m2+ 1) and x2− 4x − 2m(m2+ 1). Subtracting, we get that x− r is a factor of (2m − 4)x − (2m3− 4m2+ 2m− 4), i.e.
(2m−4)r = (2m−4)(m2+1). So m = 2 or r = m2+1. In the former case, however, both our second-degree equations become (x− 2)2= 24 and so again we have only two distinct roots. So we must have r = m2+ 1 and then substitution into (r− 2)2= 2(m3+ m + 2) gives (m2− 1)2 = 2(m3+ m + 2), which can be rewritten and factored as (m + 1)(m− 3)(m2+ 1) = 0. So m = −1 or 3; the first case has already been shown to be spurious, so we can only have m = 3.
Indeed, our equations become (x− 3)3 = 49 and (x− 2)2 = 64 so x =−6, −4, 10, and indeed we have 3 roots.
2. Let ABC be an equilateral triangle with area 7 and let M, N be points on sides AB, AC, respectively, such that AN = BM . Denote
by O the intersection of BN and CM . Assume that triangle BOC has area 2.
(a) Prove that M B/AB equals either 1/3 or 2/3.
(b) Find ∠AOB.
Solution:
(a) Let L be on BC with CL = AN , and let the intersections of CM and AL, AL and BN be P, Q, respectively. A 120-degree rotation about the center of ABC takes A to B, B to C, C to A; this same rotation then also takes M to L, L to N , N to M , and also O to P , P to Q, Q to O. Thus OP Q and M LN are equilateral triangles concentric with ABC. It follows that
∠BOC = π − ∠N OC = 2π/3, so O lies on the reflection of the circumcircle of ABC through BC. There are most two points O on this circle and inside of triangle ABC such that the ratio of the distances to BC from O and from A — i.e. the ratio of the areas of triangles OBC and ABC — can be 2/7; so once we show that M B/AB = 1/3 or 2/3 gives such positions of O it will follow that there are no other such ratios (no two points M can give the same O, since it is easily seen that as M moves along AB, O varies monotonically along its locus). If M B/AB = 1/3 then AN/AC = 1/3, and Menelaus’ theorem in triangle ABN and line CM gives BO/ON = 3/4 so [BOC]/[BN C] = BO/BN = 3/7. Then [BOC]/[ABC] = (3/7)(CN/CA) = 2/7 as desired. Similarly if M B/AB = 2/3 the theorem gives us BO/BN = 6, so [BOC]/[BN C] = BO/BN = 6/7 and [BOC]/[ABC] = (6/7)(CN/AC) = 2/7.
(b) If M B/AB = 1/3 then M ON A is a cyclic quadrilateral since
∠A = π/3 and ∠O = π − (∠P OQ) = 2π/3. Thus ∠AOB =
∠AOM + ∠M OB = ∠AN M + ∠P OQ = ∠AN M + π/3. But M B/AB = 1/3 and AN/AC = 1/3 easily give that N is the projection of M onto AC, so ∠AN M = π/2 and ∠AOB = 5π/6.
If M B/AB = 2/3 then M ON A is a cyclic quadrilateral as before, so that ∠AOB = ∠AOM +∠M OB = ∠AN M +∠P OQ.
But AM N is again a right triangle, now with right angle at M , and ∠M AN = π/3 so ∠AN M = π/6, so ∠AOB = π/2.
3. Let f (x) = x2 − 2ax − a2 − 3/4. Find all values of a such that
|f(x)| ≤ 1 for all x ∈ [0, 1].
Solution: Answer: −1/2 ≤ a ≤√
2/4. Proof: The graph of f (x) is a parabola with an absolute minimum (i.e., the leading coefficient is positive), and its vertex is (a, f (a)). Since f (0) =−a2− 3/4, we obtain that |a| ≤ 1/2 if we want f(0) ≥ −1. Now suppose a ≤ 0;
then our parabola is strictly increasing between x = 0 and x = 1 so it suffices to check f (1)≤ 1. But we have 1/2 ≤ a + 1 ≤ 1, 1/4 ≤ (a + 1)2≤ 1, 1/4 ≤ 5/4 − (a + 1)2≤ 1. Since 5/4 − (a + 1)2= f (1), we have indeed that f meets the conditions for−1/2 ≤ a ≤ 0. For a > 0, f decreases for 0≤ x ≤ a and increases for a ≤ x ≤ 1. So we must check that the minimum value f (a) is in our range, and that f (1) is in our range. This latter we get from 1 < (a + 1)2 ≤ 9/4 (since a ≤ 1/2) and so f(x) = −1 ≤ 5/4 − (a + 1)2 < 1/4. On the other hand, f (a) =−2a2− 3/4, so we must have a ≤√
2/4 for f (a)≥ −1. Conversely, by bounding f(0), f(a), f(1) we have shown that f meets the conditions for 0 < a≤√
2/4.
4. Let I and G be the incenter and centroid, respectively, of a triangle ABC with sides AB = c, BC = a, CA = b.
(a) Prove that the area of triangle CIG equals|a − b|r/6, where r is the inradius of ABC.
(b) If a = c + 1 and b = c− 1, prove that the lines IG and AB are parallel, and find the length of the segment IG.
Solution:
(a) Assume WLOG a > b. Let CM be a median and CF be the bisector of angle C; let S be the area of triangle ABC. Also let BE be the bisector of angle B; by Menelaus’ theorem on line BE and triangle ACF we get (CE/EA)(AB/BF )(F I/IC) = 1. Applying the Angle Bisector Theorem twice in triangle ABC we can rewrite this as (a/c)((a + b)/a)(F I/IC) = 1, or IC/F I = (a + b)/c, or IC/CF = (a + b)/(a + b + c). Now also by the Angle Bisector Theorem we have BF = ac/(a + b);
since BM = c/2 and a > b then M F = (a− b)c/2(a + b). So comparing triangles CM F and ABC, noting that the altitudes
to side M F (respectively AB) are equal, we have [CM F ]/S = (a− b)/2(a + b). Similarly using altitudes from M in triangles CM I and CM F (and using the ratio IC/CF found earlier), we have [CM I]/S = (a− b)/2(a + b + c); and using altitudes from I in triangles CGI and CM I gives (since CG/CM = 2/3) [CGI]/S = (a− b)/3(a + b + c). Finally S = (a + b + c)r/2 leads to [CGI] = (a− b)r/6.
(b) As noted earlier, IC/CF = (a + b)/(a + b + c) = 2/3 = CG/CM in the given case. But C, G, M are collinear, as are C, I, F , giving the desired parallelism (since line M F = line AB). We found earlier M F = (a− b)c/2(a + b) = 1/2, so GI = (2/3)(M F ) = 1/3.
5. Let n≥ 4 be an even integer and A a subset of {1, 2, . . . , n}. Consider the sums e1x1+ e2x2+ e3x3such that:
• x1, x2, x3∈ A;
• e1, e2, e3∈ {−1, 0, 1};
• at least one of e1, e2, e3 is nonzero;
• if xi= xj, then eiej6= −1.
The set A is free if all such sums are not divisible by n.
(a) Find a free set of cardinalitybn/4c.
(b) Prove that any set of cardinality bn/4c + 1 is not free.
Solution:
(a) We show that the set A ={1, 3, 5, ..., 2bn/4c − 1} is free. Any combination e1x1 + e2x2+ e3x3 with zero or two ei’s equal to 0 has an odd value and so is not divisible by n; otherwise, we have one ei equal to 0, so we have either a difference of two distinct elements of A, which has absolute value less than 2bn/4c and cannot be 0, so it is not divisible by n, or a sum (or negative sum) of two elements, in which case the absolute value must range between 2 and 4bn/4c − 2 < n and so again is not divisible by n.
(b) Suppose A is a free set; we will show |A| ≤ bn/4c. For any k, k and n− k cannot both be in A since their sum is n; likewise, n and n/2 cannot be in A. If we change any element k of A to n− k then we can verify that the set of all combinationsP eixi
taken mod n is invariant, since we can simply flip the sign of any eiassociated with the element k in any combination. Hence we may assume that A is a subset of B ={1, 2, ..., n/2 − 1}.
Let d be the smallest element of A. We group all the elements of B greater than d into “packages” of at most 2d elements, starting with the largest; i.e. we put the numbers from n/2− 2d to n/2− 1 into one package, then put the numbers from n/2− 4d to n/2 − 2d − 1 into another, and so forth, until we hit d + 1 and at that point we terminate the packaging process.
All our packages, except possibly the last, have 2d elements; so let p + 1 be the number of packages and let r be the number of elements in the last package (assume p≥ 0, since otherwise we have no packages and d = n/2− 1 so our desired conclusion holds because|A| = 1). The number of elements in B is then 2dp + r + d, so n = 4dp + 2d + 2r + 2. Note that no two elements of A can differ by d, since otherwise A is not free. Also the only element of A not in a package is d, since it is the smallest element and all higher elements of B are in packages.
Now do a case analysis on r. If r < d then each complete package has at most d elements in common with A, since the elements of any such package can be partitioned into disjoint pairs each with difference d. Thus|A| ≤ 1 + dp + r and 4|A| ≤ 4dp+4r+4≤ n (since r+1 ≤ d) so our conclusion holds. If r = d then each complete package has at most d elements in common with A, and also the last package (of d elements) has at most d− 1 elements in common with A for the following reason: its highest element is 2d, but 2d is not in A since d + d−2d = 0. So
|A| ≤ d(p+1), 4|A| < n and our conclusion holds. If r > d then we can form r− d pairs in the last package each of difference d, so each contains at most 1 element of A, and then there are 2d− r remaining elements in this package. So this package contains at most d elements, and the total number of elements in A is at least d(p + 1) + 1, so 4|A| ≤ n and our conclusion again holds.
6. Find the least natural number a for which the equation cos2π(a− x) − 2 cos π(a − x) + cos3πx
2a cosπx 2a +π
3
+ 2 = 0 has a real root.
Solution: The smallest such a is 6. The equation holds if a = 6, x = 8. To prove minimality, write the equation as
(cos π(a− x) − 1)2+ (cos(3πx/2a) cos(πx/2a + π/3) + 1) = 0;
since both terms on the left side are nonnegative, equality can only hold if both are 0. From cos π(a− x) − 1 = 0 we get that x is an integer congruent to a (mod 2). From the second term we see that each cosine involved must be−1 or 1 for the whole term to be 0; if cos(πx/2a + π/3) = 1 then πx/2a + π/3 = 2kπ for some integer k, and multiplying through by 6a/π gives 3x≡ −2a (mod 12a), while if the cosine is−1 then πx/2a + π/3 = (2k + 1)π and multiplying by 6a/π gives 3x≡ 4a (mod 12a). In both cases we have 3x divisible by 2, so x is divisible by 2 and hence so is a. Also our two cases give
−2a and 4a, respectively, are divisible by 3, so a is divisible by 3.
We conclude that 6|a and so our solution is minimal.
7. Let ABCD be a trapezoid (AB||CD) and choose F on the segment AB such that DF = CF . Let E be the intersection of AC and BD, and let O1, O2 be the circumcenters of ADF, BCF . Prove that the lines EF and O1O2 are perpendicular.
Solution: Project each of points A, B, F orthogonally onto CD to obtain A0, B0, F0; then F0 is the midpoint of CD. Also let the cir-cumcircles of AF D, BF C intersect line CD again at M, N respec-tively; then AF M D, BF N C are isosceles trapezoids and F0M = DA0, N F0 = B0C. Let x = DA0, y = A0F0 = AF , z = F0B0 = F B, w = B0C, using signed distances throughout (x < 0 if D is be-tween A0and F0, etc.), so we have x + y = z + w; call this value S, so DC = 2S. Also let line F E meet DC at G; since a homothety about E with (negative) ratio CD/AB takes triangle ABE into CDE it also takes F into G, so DG/GC = F B/AF = F0B0/A0B0 = z/y and we easily get DG = 2zS/(y + z), GC = 2yS/(y + z). Now
N F0 = w, DF0= S implies DN = z and so DN/DG = (y + z)/2S.
Similarly F0M = x, F0C = S so M C = y and M C/GC = (y + z)/2S also. So DN/DG = M C/GC, N G/DG = GM/GC and N G· GC = DG· GM. Since NC and DM are the respective chords of the circumcircles of BF C and ADC that contain point G we conclude that G has equal powers with respect to these two circles, i.e. it is on the radical axis. F is also on the axis since it is an intersection point of the circles, so the line F GE is the radical axis, which is perpendicular to the line O1O2connecting the centers of the circles.
8. Find all natural numbers n for which a convex n-gon can be di-vided into triangles by diagonals with disjoint interiors, such that each vertex of the n-gon is the endpoint of an even number of the diagonals.
Solution: We claim that 3|n is a necessary and sufficient condition.
To prove sufficiency, we use induction of step 3. Certainly for n = 3 we have the trivial dissection (no diagonals drawn). If n > 3 and 3|n then let A1, A2, . . . , An be the vertices of an n-gon in counterclock-wise order; then draw the diagonals A1An−3, An−3An−1, An−1A1; these three diagonals divide our polygon into three triangles and an (n− 3)-gon A1A2. . . An−3. By the inductive hypothesis the latter can be dissected into triangles with evenly many diagonals at each vertex, so we obtain the desired dissection of our n-gon, since each vertex from A2 through An−4 has the same number of diagonals in the n-gon as in the (n− 3)-gon (an even number), A1and An−3each have two diagonals more than in the (n− 3)-gon, while An−1 has 2 diagonals and An and An−2 have 0 each.
To show necessity, suppose we have such a decomposition of a poly-gon with vertices A1, A2, . . . , An in counterclockwise order, and for convenience assume labels are mod n. Call a diagonal AiAj in our dissection a “right diagonal” from Aiif no point Ai+2, Ai+3, . . . , Aj−1
is joined to Ai (we can omit Ai+1 from our list since it is joined by an edge). Clearly every point from which at least one diagonal emanates has a unique right diagonal. Also we have an important lemma: if AiAj is a right diagonal from Ai, then within the polygon AiAi+1. . . Aj, each vertex belongs to an even number of diagonals.
Proof: Each vertex from any of the points Ai+1, . . . , Aj−1 belongs to an even number of diagonals of the n-gon, but since the diagonals
of the n-gon are nonintersecting these diagonals must lie within our smaller polygon, so we have an even number of such diagonals for each of these points. By hypothesis, Aiis not connected via a diag-onal to any other point of this polygon, so we have 0 diagdiag-onals from Ai, an even number. Finally evenly many diagonals inside this poly-gon stem from Aj, since otherwise we would have an odd number of total endpoints of all diagonals.
Now we can show 3|n by strong induction on n. If n = 1 or 2, then there is clearly no decomposition, while if n = 3 we have 3|n. For n > 3 choose a vertex Ai1 with some diagonal emanating from it, and let Ai1Ai2 be the right diagonal from Ai1. By the lemma there are evenly many diagonals from Ai2 with their other endpoints in {Ai1+1, Ai1+2, . . . , Ai2−1}, and one diagonal Ai1Ai2, so there must be at least one other diagonal from Ai2 (since the total number of di-agonals there is even). This implies Ai1Ai2 is not the right diagonal from Ai2, so choose the right diagonal Ai2Ai3. Along the same lines we can choose the right diagonal Ai3Ai4 from Ai3, with Ai2 and Ai4
distinct, then continue with Ai4Ai5 as the right diagonal from Ai4, etc. Since the diagonals of the n-gon are nonintersecting this pro-cess must terminate with some Aik+1 = Ai1. Now examine each of the polygons AixAix+1Aix+2. . . Aix+1, x = 1, 2, . . . , k (indices x are taken mod k). By the lemma each of these polygons is divided into triangles by nonintersecting diagonals with evenly many diagonals at each vertex, so by the inductive hypothesis the number of vertices of each such polygon is divisible by 3. Also consider the polygon Ai1Ai2. . . Aik. We claim that in this polygon, each vertex belongs to an even number of diagonals. Indeed, from Aix we have an even number of diagonals to points in Aix−1+1, Aix−1+2, . . . , Aix−1, plus the two diagonals Aix−1Aix and AixAix+1. This leaves an even num-ber of diagonals from Aix to other points; since Aix was chosen as the endpoint of a right diagonal we have no diagonals lead to points in Aix+1, . . . , Aix+1−1, so it follows from the nonintersecting crite-rion that all remaining diagonals must lead to points Aiy for some y.
Thus we have an even number of diagonals from Aix to points Aiy for some fixed x; it follows from the induction hypothesis that 3|k.
So, if we count each vertex of each polygon AixAix+1Aix+2. . . Aix+1 once and then subtract the vertices of Ai1Ai2. . . Aik, each vertex of our n-gon is counted exactly once, but from the above we have been adding and subtracting multiples of 3. Thus we have 3|n.
9. For any real number b, let f (b) denote the maximum of the function
sin x + 2 3 + sin x+ b
over all x∈ R. Find the minimum of f(b) over all b ∈ R.
Solution: The minimum value is 3/4. Let y = 3 + sin x; note y ∈ [2, 4] and assumes all values therein. Also let g(y) = y + 2/y;
this function is increasing on [2, 4], so g(2) ≤ g(y) ≤ g(4). Thus 3 ≤ g(y) ≤ 9/2, and both extreme values are attained. It now fol-lows that the minimum of f (b) = max(|g(y) + b − 3|) is 3/4, which is attained by b = −3/4; for if b > −3/4 then choose x = π/2 so y = 4 and then g(y) + b− 3 > 3/4, while if b < −3/4 then choose x =−π/2 so y = 2 and g(y) + b − 3 = −3/4; on the other hand, our range for g(y) guarantees−3/4 ≤ g(y) + b − 3 ≤ 3/4 for b = −3/4.
10. Let ABCD be a convex quadrilateral such that ∠DAB = ∠ABC =
∠BCD. Let H and O denote the orthocenter and circumcenter of the triangle ABC. Prove that H, O, D are collinear.
Solution: Let M be the midpoint of B and N the midpoint of BC. Let E = AB∩ CD and F = BC ∩ AD. Then EBC and F AB are isosceles triangles, so EN∩ F M = 0. Thus applying Pappus’s theorem to hexagon M CEN AF , we find that G, O, D are collinear, so D lies on the Euler line of ABC and H, O, D are collinear.
11. For any natural number n≥ 3, let m(n) denote the maximum num-ber of points lying within or on the boundary of a regular n-gon of side length 1 such that the distance between any two of the points is greater than 1. Find all n such that m(n) = n− 1.
Solution: The desired n are 4, 5, 6. We can easily show that m(3) = 1, e.g. dissect an equilateral triangle ABC into 4 congruent triangles and then for two points P, Q there is some corner triangle inside which neither lies; if we assume this corner is at A then the circle with diameter BC contains the other three small triangles and so contains P and Q; BC = 1 so P Q ≤ 1. This method will be useful later; call it a lemma.
On the other hand, m(n)≥ n − 1 for n ≥ 4 as the following process indicates. Let the vertices of our n-gon be A1, A2, . . . , An. Take P1 = A1. Take P2 on the segment A2A3 at an extremely small distance d2from A2; then P2P1> 1, as can be shown rigorously, e.g.
using the Law of Cosines in triangle P1A2P2 and the fact that the cosine of the angle at A2 is nonnegative (since n≥ 4). Moreover P2
is on a side of the n-gon other than A3A4, and it is easy to see that as long as n≥ 4, the circle of radius 1 centered at A4 intersects no side of the n-gon not terminating at A4, so P2A4 > 1 while clearly P2A3< 1. So by continuity there is a point P3on the side A3A4with P2P3= 1. Now slide P3by a small distance d3on A3A4towards A4; another trigonometric argument can easily show that then P2P3> 1.
Continuing in this manner, obtain P4 on A4A5 with P3P4 = 1 and slide P4 by distance d4 so that now P3P4 > 1, etc. Continue doing this until all points Pi have been defined; distances PiPi+1 are now greater than by construction, Pn−1P1 > 1 because P1 = A1 while Pn−1 is in the interior of the side An−1An; and all other PiPj are greater than 1 because it is easy to see that the distance between any two points of nonadjacent sides of the n-gon is at least 1 with equality possible only when (among other conditions) Pi, Pjare endpoints of their respective sides, and in our construction this never occurs for distinct i, j. So our construction succeeds. Moreover, as all the distances di tend to 0 each Pi tends toward Ai, so it follows that the maximum of the distances AiPican be made as small as desired by choosing di sufficiently small. On the other hand, when n > 6 the center O of the n-gon is at a distance greater than 1 from each vertex, so if the Pi are sufficiently close to the Ai then we will also have OPi > 1 for each i. Thus we can add the point O to our set, showing that m(n)≥ n for n > 6.
It now remains to show that we cannot have more than n− 1 points at mutual distances greater than 1 for n = 4, 5, 6. As before let the vertices of the polygon be A1, etc. and the center O; suppose we have n points P1, . . . , Pn with PiPj > 1 for i not equal to j. Since n≤ 6 it follows that the circumradius of the polygon is not greater than 1, so certainly no Pican be equal to O. Let the ray from O through Pi intersect the polygon at Qiand assume WLOG our numbering is such that Q1, Q2, . . . , Qnoccur in that order around the polygon, in the same orientation as the vertices were numbered. Let Q1be on the side AkAk+1. A rotation by angle 2π/n brings Ak into Ak+1; let it
also bring Q1into Q01, so triangles Q1Q01O and AkAk+1O are similar.
We claim P2 cannot lie inside or on the boundary of quadrilateral
We claim P2 cannot lie inside or on the boundary of quadrilateral