Split (1)

train · 617k rows

text stringlengths 2 132k	source dict
1 “ f 4pn ` 1q. (2) I. Let us denote by Ri the range of f i; note that R0 “ Zě0 since f 0 is the identity function. Obviously, R0 Ě R1 Ě . . . . Next, from (2) we get that if a P R4 then also a ` 1 P R4. This implies that Zě0zR4 — and hence Zě0zR1 — is finite. In particular, R1 is unbounded. Assume that f pmq “ f pnq f...	{ "page_id": null, "source": 6828, "title": "from dpo" }
S1. Otherwise b ´ 1 P Zě0, and there exists some n ą 0 such that f pnq “ b ´ 1; but then f 3pn ´ 1q “ f pnq ` 1 “ b,so b P R3.This yields 3k “ \| S1 Y S2 Y S3\| ď 1 ` 1 ` \| S1\| “ k ` 2, or k ď 1. Therefore k “ 1, and the inequality above comes to equality. So we have S1 “ t au, S2 “ t f paqu , and S3 “ t f 2paqu for some...	{ "page_id": null, "source": 6828, "title": "from dpo" }
“ f 2pn ´ 1q “ f pnq, establishing the step. In this case we have obtained the first of two answers; checking that is satisfies p˚q is straightforward. Case 2. Assume now that f 2paq “ 0; then by (3) we get a “ f p0q ` 1. By (4) we get f pa ` 1q “ f 2paq “ 0, then f p0q “ f 3paq “ f pa ` 1q ` 1 “ 1, hence a “ f p0q ` 1...	{ "page_id": null, "source": 6828, "title": "from dpo" }
1q ` 1, f 3p4k ` 1q “ 4k ` 4 “ f p4k ` 2q ` 1,f 3p4k ` 2q “ 4k ` 1 “ f p4k ` 3q ` 1, f 3p4k ` 3q “ 4k ` 6 “ f p4k ` 4q ` 1. Solution 2. I. For convenience, let us introduce the function gpnq “ f pnq ` 1. Substituting f pnq instead of n into p˚q we obtain f 4pnq “ f `f pnq ` 1˘ ` 1, or f 4pnq “ g2pnq. (5) Applying f to ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
pmod cq. Thus, m ” n pmod cq ðñ f pmq ” f pnq pmod cq ðñ gpmq ” gpnq pmod cq. (9) Now, let us introduce the function δpnq “ f pnq ´ n “ gpnq ´ n ´ 1. Set S “ > c´1 ÿ > n“0 δpnq. Using (8), we get that for every complete residue system n1, . . . , n c modulo c we also have S “ > c ÿ > i“1 δpniq. By (9), we get that tf k...	{ "page_id": null, "source": 6828, "title": "from dpo" }
3u. If d “ 1 then we have f p0q “ gp0q ´ 1 “ 0 which is impossible since f pnq ‰ n for all n. If d “ 3 then gp3q “ g2p0q “ 4and hence f p3q “ 3 which is also impossible. Thus gp0q “ 2 and hence gp2q “ g2p0q “ 4. Next, if gp1q “ 1 ` 4k for some integer k, then 5 “ g2p1q “ gp1 ` 4kq “ gp1q ` 4k “ 1 ` 8k which is impossib...	{ "page_id": null, "source": 6828, "title": "from dpo" }
xn`1 on both sides gives anpn ´ 2mqp n ´ 1q “ 0, so n “ 1 or n “ 2m.If n “ 1, one easily verifies that P pxq “ x is a solution, while P pxq “ 1 is not. Since the given condition is linear in P , this means that the linear solutions are precisely P pxq “ tx for t P R.Now assume that n “ 2m. The polynomial xP px ` 1q ´ p...	{ "page_id": null, "source": 6828, "title": "from dpo" }
´ 1, r ` b are roots of S,while r ´ a ´ 1 and r ` b ` 1 are not. The two statements above imply that r ´ a is a root of B and r ` b is a root of A.Since r ´ a is a root of Bpxq and of Apx ` a ` bq, it is also a root of their greatest common divisor Cpxq as integer polynomials. If Cpxq was a non-trivial divisor of Bpxq,...	{ "page_id": null, "source": 6828, "title": "from dpo" }
RQ. Each root is taken with its multiplicity. Then the multiset of complex roots of Qpx ´ 1q is RQ ` 1 “ t z ` 1 : z P RQu.20 IMO 2013 Colombia Let tx1, x 2, x 3u and ty1, y 2, y 3u be the multisets of roots of the polynomials Apxq “ x3 ´ mx 2 ` 1and Bpxq “ x3 ` mx 2 ` 1, respectively. From (2) we get the equality of m...	{ "page_id": null, "source": 6828, "title": "from dpo" }
and one if a ě ´ 1. Now, since yi ´ xi P Z for i “ 1, 2, 3, the cubics Am and A´m must have the same number of real roots. The previous analysis then implies that m “ 1 or m “ ´ 1. Therefore the real root α of A1pxq “ x3 ` x2 ` 1 and the real root β of A´1pxq “ x3 ´ x2 ` 1 must differ by an integer. But this is impossi...	{ "page_id": null, "source": 6828, "title": "from dpo" }
1. Therefore k ě 2n ´ 1. It remains to show that a suitable partition into 2 n ´ 1 groups always exists. We proceed by induction on d. For d ď 2n ´ 1 the result is trivial. If d ě 2n, then since pa1 ` a2q ` . . . ` p a2n´1 ` a2nq ď n we may find two numbers ai, a i`1 such that ai ` ai`1 ď 1. We “merge” these two number...	{ "page_id": null, "source": 6828, "title": "from dpo" }
. . . , a du. In the example above, this partition is ta1, a 2u, ta3u, ta4, a 5u, ta6u, H, ta7u, ta8, a 9, a 10 u. We claim that in this partition, the sum of the numbers in this group is at most 1. For the sets tait u this is obvious since ait ď 1. For the sets tait ` 1, . . . , a it`1´1u this follows from the fact th...	{ "page_id": null, "source": 6828, "title": "from dpo" }
real numbers between 0 and 1 whose sum is a real number r. In this case the smallest value of k is k “ r2rs ´ 1, as Solution 3 shows. Solutions 1 and 2 lead to the slightly weaker bound k ď 2rrs ´ 1. This is actually the optimal bound for partitions into consecutive groups, which are the ones contemplated in these two ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
colors. Thus, if the goal is reached, then each arc should intersect some of the drawn lines. Since any line contains at most two points of the circle, one needs at least 4026 {2 “ 2013 lines. It remains to prove that one can reach the goal using 2013 lines. First of all, let us mention that for every two points A and ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
both are blue, and one may act as in Case 2. Solution 2. Let us present a different proof of the fact that k “ 2013 suffices. In fact, we will prove a more general statement: If n points in the plane, no three of which are collinear, are colored in red and blue arbitrarily, then it suffices to draw tn{2u lines to reach...	{ "page_id": null, "source": 6828, "title": "from dpo" }
induction is proved. > Comment 2. One may ask a more general question, replacing the numbers 2013 and 2014 by any positive integers mand n, say with mďn. Denote the answer for this problem by fpm, n q.One may show along the lines of Solution 1 that mďfpm, n q ď m`1; moreover, if mis even then > fpm, n q “ m.On the othe...	{ "page_id": null, "source": 6828, "title": "from dpo" }
are entangled. Recall that a proper coloring of a graph G is a coloring of its vertices in several colors so that every two connected vertices have different colors. Lemma. Assume that a graph G admits a proper coloring in n colors ( n ą 1). Then one may perform a sequence of operations resulting in a graph which admit...	{ "page_id": null, "source": 6828, "title": "from dpo" }
obtain a graph admitting a proper coloring in one color, that is — a graph with no edges, as required. Solution 2. Again, we will use the graph language. I. We start with the following observation. Lemma. Assume that a graph G contains an isolated vertex A, and a graph G˝ is obtained from G by deleting this vertex. The...	{ "page_id": null, "source": 6828, "title": "from dpo" }
where A0 > i is the common “ancestor” of all the vertices Aji , and the binary expansion of j (adjoined with some zeroes at the left to have m digits) “keeps the history” of this vertex: the dth digit from the right is 0 if at the dth doubling the ancestor of Aji was in the original part, and this digit is 1 if it was ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
all degrees being odd. In this graph, we delete one by one all the vertices Ajn where the sum of the binary digits of j is even; it is possible since there are no edges between such vertices, so all their degrees remain odd. After that, we delete all isolated vertices. Step 3. Finally, consider any remaining vertex Ajn...	{ "page_id": null, "source": 6828, "title": "from dpo" }
The number of different parts in such a partition is the number of (distinct) elements in the set ta1, a 2, . . . , a ku.We say that an A-partition of n into k parts is optimal if there is no A-partition of n into r parts with r ă k. Prove that any optimal A-partition of n contains at most 3 ?6n different parts. (Germa...	{ "page_id": null, "source": 6828, "title": "from dpo" }
represent the elements of S by a sequence of dots in increasing order, and represent a subset of S by shading in the appropriate dots, we have: Ski,j “ ‚ ‚ ‚ ‚ ‚ ‚ ‚ looooomooooon > k´i ˝ ˝ ˝ ˝ ˝ looomooon > j ‚ ˝ ˝ ˝ ˝ ˝ ˝ ˝ looooomooooon > s´k´j ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ looooooomooooooon > i´1 Denote by Σ ki,j the sum of el...	{ "page_id": null, "source": 6828, "title": "from dpo" }
s3 6 ą n. Since they are between 1 and n, at least two of these integers are equal. Consequently, there exist 1 ď k ă k1 ď s and X “ Ski,j as well as Y “ Sk1 > i1,j 1 such that ÿ > xPX x “ ÿ > yPY y, but \|X\| “ k ă k1 “ \| Y \|, as required. The result follows. 28 IMO 2013 Colombia Solution 2. Assume, to the contrary, tha...	{ "page_id": null, "source": 6828, "title": "from dpo" }
> 2 .Set b0 “ 0, Σ 0,0 “ 0, and Σ i,j “ b1`¨ ¨ ¨` bi´1`bj for 1 ď i ď j ă s. There are sps´1q > 2 `1 ą bs such sums; so at least two of them, say Σ i,j and Σ i1,j 1 , are congruent modulo bs (where pi, j q ‰ p i1, j 1q). This means that Σ i,j ´ Σi1,j 1 “ rb s for some integer r. Notice that for i ď j ă k ă s we have 0 ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
proof here. Let k “ t 3 ?2nu ´ 1. The statement is trivial for n ă 4, so we assume n ě 4 and hence k ě 1. Let h “ t n´1 > k u. Notice that h ě n > k ´ 1. Now let A “ t 1, . . . , h u, and set a1 “ h, a2 “ h ´ 1, . . . , ak “ h ´ k ` 1, and ak`1 “ n ´ p a1 ` ¨ ¨ ¨ ` akq.It is not difficult to prove that ak ą ak`1 ě 1, w...	{ "page_id": null, "source": 6828, "title": "from dpo" }
sequence. Assume that for every nonnegative integer m there exists a nonnegative integer n P r m ` 1, m ` rs such that bm “ bn. Then for every indices k ď ℓ there exists an index t P r ℓ, ℓ ` r ´ 1s such that bt “ bk . Moreover, there are at most r distinct numbers among the terms of pbiq. Proof. To prove the first cla...	{ "page_id": null, "source": 6828, "title": "from dpo" }
k ě 0 two of the tuples Ak, A k`1, . . . , A k`rr are identical. This means that there exists a positive integer p ď rr such that the equality Ad “ Ad`p holds infinitely many times. Let D be the set of indices d satisfying this relation. Now we claim that D coincides with the set of all nonnegative integers. Since D is...	{ "page_id": null, "source": 6828, "title": "from dpo" }
The conclusion remains true even if the problem condition only holds for every s ě N for some positive integer N . To show that, one can act as follows. Firstly, the sums of the form Spi, i ` N q attain at most r values, as well as the sums of the form Spi, i `N `1q. Thus the terms ai “ Spi, i ` N ` 1q´ Spi ` 1, i ` N ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
q “ iu the set of cities at distance exactly i from city a.Assume that for some city x the set D “ S4pxq has size at least 2551. Let A “ S1pxq. Asubset A1 of A is said to be substantial , if every city in D can be reached from x with four flights while passing through some member of A1; in other terms, every city in D ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
that all 2 pm ´ 1q cities of the form by , cy with y P Aa are distinct. Indeed, no by may coincide with any cz since their distances from x are different. On the other hand, if one had by “ bz for y ‰ z, then there would exist a path of length 4 from x to dz via y, namely x–y–bz –cz–dz ; this is impossible by the choic...	{ "page_id": null, "source": 6828, "title": "from dpo" }
the statement of the problem may be replaced by n and Y pn`1q2 > 4 ] for any positive integer n. Still more generally, one can also replace the pair p3, 4q of distances under consideration by any pair pr, s q of positive integers satisfying r ă s ď 3 > 2 r.To adapt the above proof to this situation, one takes A “ Ss´rp...	{ "page_id": null, "source": 6828, "title": "from dpo" }
` d, the chord joining numbers a and c does not intersect the chord joining numbers b and d.Let M be the number of beautiful arrangements of 0 , 1, . . . , n . Let N be the number of pairs px, y q of positive integers such that x ` y ď n and gcd px, y q “ 1. Prove that M “ N ` 1. (Russia) Solution 1. Given a circular a...	{ "page_id": null, "source": 6828, "title": "from dpo" }
the endpoints of these segments. If x and y are their respective partners, we have n ě 0 ` x “ k “ n ` y ě n. Thus 0 and n are the endpoints of one of the chords; say it is C.Let D be the chord formed by the numbers u and v which are adjacent to 0 and n and on the same side of C as A and B, as shown in Figure 2. Set t ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
Type 2 otherwise; i.e. , if 0 is aligned with these n–chords. We will show that each Type 1 arrangement of r0, n ´ 1s arises from a unique arrangement of r0, n s, and each Type 2 arrangement of r0, n ´ 1s arises from exactly two beautiful arrangements of r0, n s.If T is of Type 1, let 0 lie between chords A and B. Sinc...	{ "page_id": null, "source": 6828, "title": "from dpo" }
remains to show that Ln´1 is the number of pairs px, y q of positive integers with x ` y “ n and gcd px, y q “ 1. Since n ě 3, this number equals ϕpnq “ #tx : 1 ď x ď n, gcd px, n q “ 1u.To prove this, consider a Type 2 beautiful arrangement of r0, n ´ 1s. Label the positions 0, . . . , n ´ 1 pmod nq clockwise around t...	{ "page_id": null, "source": 6828, "title": "from dpo" }
az q, which means that the chord from w to y and the chord from x to z are parallel. Thus ( 1) is beautiful, and by construction it has Type 2. The desired result follows. Shortlisted problems – solutions 35 Solution 2. Notice that there are exactly N irreducible fractions f1 ă ¨ ¨ ¨ ă fN in p0, 1q whose denominator is...	{ "page_id": null, "source": 6828, "title": "from dpo" }
b to d in Apαq. Hence in a cyclic arrangement all k—chords are parallel. In particular every cyclic arrangement is beautiful. Next we show that there are exactly N ` 1 distinct cyclic arrangements. To see this, let us see how Apαq changes as we increase α from 0 to 1. The order of points p and q changes precisely when ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
later: if fi ă α ă fi`1 then 0 is immediately after bi`1 and before bi in Apαq. (2) Indeed, we already observed that bi is the first number after 0 in Apfi ` ǫq “ Apαq. Similarly we see that bi`1 is the last number before 0 in Apfi`1 ´ ǫq “ Apαq.36 IMO 2013 Colombia Finally, we show that any beautiful arrangement of r0...	{ "page_id": null, "source": 6828, "title": "from dpo" }
clockwise order, as shown in Figure 4. Similarly, n ´ 1 is between y and y ´ 1. Therefore x, y, x ´ 1, n ´ 1, and y ´ 1 occur in this order in Anpαq and hence in A (possibly with y “ x ´ 1 or x “ y ´ 1). Now, A may only differ from Anpαq in the location of n. In A, since the chord from n ´ 1to x and the chord from n to...	{ "page_id": null, "source": 6828, "title": "from dpo" }
on the real line. Player A has a pot of paint with four units of black ink. A quantity p of this ink suffices to blacken a (closed) real interval of length p. In every round, player A picks some positive integer m and provides 1 {2m units of ink from the pot. Player B then picks an integer k and blackens the interval f...	{ "page_id": null, "source": 6828, "title": "from dpo" }
make the convention that, at the beginning of the game, the interval r1, 2s is already blackened; thus, if yr ` 1 “ 2m, then B blackens Ir > 0 .Our aim is to estimate the amount of ink used after each round. Firstly, we will prove by induction that, if before rth round the segment r0, 1s is not completely colored, then...	{ "page_id": null, "source": 6828, "title": "from dpo" }
ą 1{2m.Next, any interval blackened by B before the rth move which intersects pxr, x r`1q should be contained in rxr, x r`1s; by pii q, all such intervals have different lengths not exceeding 1 {2m, so the total amount of ink used for them is less than 2 {2m. Thus, the amount of ink used for the segment r0, x r`1s does...	{ "page_id": null, "source": 6828, "title": "from dpo" }
of the type r0, x s. Such a strategy cannot work (even if there is more ink available). Indeed, under the assumption that B uses such a strategy, let us prove by induction on s the following statement: For any positive integer s, player A has a strategy picking only positive integers m ď s in which, if player B ever pa...	{ "page_id": null, "source": 6828, "title": "from dpo" }
ink suffices to blacken a (closed) real interval of length p. In the beginning of the game, player A chooses (and announces) a positive integer N . In every round, player A picks some positive integer m ď N and provides 1{2m units of ink from the pot. The player B picks an integer k and blackens the interval from k{2m ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
W Z is the radical axis of ω1 and ω2; similarly, BN is the radical axis of ω1 and ω3, and CM is the radical axis of ω2 and ω3. Hence A “ BN X CM is the radical center of the three circles, and therefore W Z passes through A.Since W X and W Y are diameters in ω1 and ω2, respectively, we have =W ZX “ =W ZY “ 90 ˝,so the ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
not containing A.The circumcircles of the triangles AMT and ANT intersect the perpendicular bisectors of AC and AB at points X and Y , respectively; assume that X and Y lie inside the triangle ABC . The lines MN and XY intersect at K. Prove that KA “ KT . (Iran) Solution 1. Let O be the center of ω, thus O “ MY X NX . ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
M and CT L are congruent, so CL “ BM “ AM .Let X1 be the common point of the line NX with the external bisector of =BAC ; notice that it lies outside the triangle ABC . Then we have =T AX 1 “ 90 ˝ and X1A “ X1C, so we get =X1AM “ 90 ˝ ` =BAC {2 “ 180 ˝ ´ =X1AC “ 180 ˝ ´ =X1CA “ =X1CL . Thus the triangles X1AM and X1CL ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
N and X1Y 1 (which is the external bisector of =BAC )do not intersect on the perpendicular bisector of AT . Thus, any solution should involve some argument using the choice of the intersection points X and Y .42 IMO 2013 Colombia ## G3. In a triangle ABC , let D and E be the feet of the angle bisectors of angles A and ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
AB q “ dpL, AB q ` dpN, AB q “ dpK, AB q ` dpM, AB q “ apsin δ ` sin εq. Thus the condition (1) reads sin μ ` sin ν “ sin δ ` sin ε. (2) > αβδεψψμνAB > C > D > E > K > L > M > N If one of the angles α and β is non-acute, then the desired inequality is trivial. So we assume that α, β ă π{2. It suffices to show then that...	{ "page_id": null, "source": 6828, "title": "from dpo" }
let =ACB “ γ. Note that the condition γ ă =CBA implies γ ă 90 ˝. Since =P BA “ γ, the line P B is tangent to ω, so P A ¨ P C “ P B 2 “ P D 2. By P A > P D “ P D > P C the triangles P AD and P DC are similar, and =ADP “ =DCP .Next, since =ABQ “ =ACB , the triangles ABC and AQB are also similar. Then =AQB “ =ABC “ =ARC ,...	{ "page_id": null, "source": 6828, "title": "from dpo" }
and V AE are also similar. Let P H be an altitude in the isosceles triangle P BD ; then BH “ HD . Let G be the intersection point of P H and AB . By the symmetry with respect to P H , we have =BDG “ =DBG “ γ “ =BEA ; thus DG ‖ AE and hence BG > GA “ BD > DE . Thus the points G and D correspond to each other in the simi...	{ "page_id": null, "source": 6828, "title": "from dpo" }
the second point of intersection of ω and BD . Due to the isosceles triangle BDP ,the tangent of ω at E is parallel to DP and consequently it intersects BP at some point L. Of course, P D ‖ LE . Let M be the midpoint of BE , and let H be the midpoint of BR . Notice that =AEB “ =ACB “ =ABQ “ =ABE , so A lies on the perp...	{ "page_id": null, "source": 6828, "title": "from dpo" }
QA > ωBAUOXQEZLPY Figure 3 Figure 4 In order to show a third fixed point, draw a line parallel with LE through A; let that line intersect BE , LB and ω at X, Y and Z ‰ A, respectively (see Fig. 4). We show that X is a fixed point. The images of X at the first three transformations are X Þ Ñ Y Þ Ñ X Þ Ñ Z. From =XBZ “ =...	{ "page_id": null, "source": 6828, "title": "from dpo" }
z “ EF “ BC . Consider the points P, Q, and R such that the quadrilaterals CDEP , EF AQ , and ABCR are parallelograms. We compute =P EQ “ =F EQ ` =DEP ´ =E “ p 180 ˝ ´ =F q ` p 180 ˝ ´ =Dq ´ =E “ 360 ˝ ´ =D ´ =E ´ =F “ 1 > 2 `=A ` =B ` =C ´ =D ´ =E ´ =F ˘ “ θ{2. Similarly, =QAR “ =RCP “ θ{2. > D > E > F > A > B > CP > ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
perpendicular bisector of P Q and AD is the perpendicular bisector of QR . It follows that AD , BE , and CF are concurrent at the circumcenter of P QR .Shortlisted problems – solutions 47 Solution 2. Let X “ CD X EF , Y “ EF X AB , Z “ AB X CD , X1 “ F A X BC , Y 1 “ BC X DE , and Z1 “ DE X F A . From =A ` =B ` =C “ 36...	{ "page_id": null, "source": 6828, "title": "from dpo" }
ÝÝÑ EF “ ÝÑ 0 , or else we would have two vectors with different directions whose sum is ÝÑ 0 . > T > DEF > A > B > C > Z > X > Y > Z′ > X′ > Y′ > O3 > O1 > O2 > N > LMO This allows us to consider a triangle LMN with ÝÝÑ LM “ ÝÝÑ EF , ÝÝÑ MN “ ÝÝÑ AB , and ÝÝÑ NL “ ÝÝÑ CD . Let O be the circumcenter of △LMN and conside...	{ "page_id": null, "source": 6828, "title": "from dpo" }
2 and F O 3 are translations of NO and MO respectively, this implies >pT O 3, T O 2q “ >pCO 2, F O 3q. (3) 48 IMO 2013 Colombia Adding the three equations in ( 2) and ( 3) gives >pT F, T C q “ >pCT, F T q “ ´ >pT F, T C q which implies that T is on CF . Analogous arguments show that it is on AD and BE also. The desired...	{ "page_id": null, "source": 6828, "title": "from dpo" }
Therefore A “ 0, B “ a, C “ a ` cr, D “ cpr ´ 1q, E “ ´ br ´ c, F “ ´ br. Now consider a point W on AD given by the complex number cpr ´ 1qλ, where λ is a real number with 0 ă λ ă 1. Since D ‰ A, we have r ‰ 1, so we can define s “ 1{p r ´ 1q. From rr “ \| r\|2 “ 1we get 1 ` s “ r r ´ 1 “ r r ´ rr “ 1 1 ´ r “ ´ s. Now, W...	{ "page_id": null, "source": 6828, "title": "from dpo" }
of the triangle ABC . Prove that the triangle ABC is right-angled. (Russia) Solution 1. Denote the circumcircles of the triangles ABC and A1B1C1 by Ω and Γ, respectively. Denote the midpoint of the arc CB of Ω containing A by A0, and define B0 as well as C0 analogously. By our hypothesis the centre Q of Γ lies on Ω. Le...	{ "page_id": null, "source": 6828, "title": "from dpo" }
Q and B1 lie on different sides of A1C1; obviously, the same holds for the points B and B1. So, the points Q and B are on the same side of A1C1.Notice that the perpendicular bisector of A1C1 intersects Ω at two points lying on different sides of A1C1. By the first statement from the Lemma, both points B0 and Q are amon...	{ "page_id": null, "source": 6828, "title": "from dpo" }
Then BC 1 “ BA 1 and consequently the triangle ABC is isosceles at B. Moreover we have BC 1 “ B1C in any triangle, and hence BB 1 “ BC 1 “ B1C;similarly, BB 1 “ B1A. It follows that B1 is the centre of Ω and that the triangle ABC has a right angle at B.So from now on we may suppose Q R t A, B, C u. We start with the fo...	{ "page_id": null, "source": 6828, "title": "from dpo" }
we may suppose that Q lies in the interior of the arc AB of Ω not containing C. We will sometimes use tacitly that the six trian-gles QBA 1, QA 1C, QCB 1, QB 1A, QC 1A, and QBC 1 have the same orientation. As Q cannot be the circumcentre of the triangle ABC , it is impossible that QA “ QB “ QC and thus we may also supp...	{ "page_id": null, "source": 6828, "title": "from dpo" }
“ =B1QA. (3) From (1) and (3) we get p=B1QC ` =B1QA q ` p =C1QB ´ =BQA 1q “ 180 ˝, i.e. =CQA ` =A1QC 1 “ 180 ˝. In light of (2) this may be rewritten as 2 =CQA “ 180 ˝ and as Q lies on Ω this implies that the triangle ABC has a right angle at B. A B C A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1...	{ "page_id": null, "source": 6828, "title": "from dpo" }
n for all n ě 2. This is trivially true for n “ 1 also. It follows that f pnq “ n for all n. This function obviously satisfies the desired property. Solution 2. Setting m “ f pnq we get f pnqp f pnq` 1q \| f pnqf pf pnqq` n. This implies that f pnq \| n for all n.Now let m be any positive integer, and let p ą 2m2 be a pr...	{ "page_id": null, "source": 6828, "title": "from dpo" }
m k such that 1 ` 2k ´ 1 n “ ˆ 1 ` 1 m1 ˙ ˆ 1 ` 1 m2 ˙ ¨ ¨ ¨ ˆ 1 ` 1 mk ˙ . (Japan) Solution 1. We proceed by induction on k. For k “ 1 the statement is trivial. Assuming we have proved it for k “ j ´ 1, we now prove it for k “ j. Case 1. n “ 2t ´ 1 for some positive integer t.Observe that 1 ` 2j ´ 1 2t ´ 1 “ 2pt ` 2j´...	{ "page_id": null, "source": 6828, "title": "from dpo" }
´ 1 and ´n modulo 2 k: n ´ 1 ” 2a1 ` 2a2 ` ¨ ¨ ¨ ` 2ar pmod 2 kq where 0 ď a1 ă a2 ă . . . ă ar ď k ´ 1, ´n ” 2b1 ` 2b2 ` ¨ ¨ ¨ ` 2bs pmod 2 kq where 0 ď b1 ă b2 ă . . . ă bs ď k ´ 1. Since ´1 ” 20 ` 21 ` ¨ ¨ ¨ ` 2k´1 pmod 2 kq, we have ta1, . . . , a ru Y t b1 . . . , b su “ t 0, 1, . . . , k ´ 1u and r ` s “ k. Write...	{ "page_id": null, "source": 6828, "title": "from dpo" }
ď s, the desired equality holds. It remains to check that every mi is an integer. For 1 ď p ď r we have n ` Sp`1 ` Ts ” n ` Ts ” 0 pmod 2 ap q and for 1 ď q ď r we have n ` Tq´1 ” n ` Ts ” 0 pmod 2 bq q. The desired result follows. Shortlisted problems – solutions 55 ## N3. Prove that there exist infinitely many positi...	{ "page_id": null, "source": 6828, "title": "from dpo" }
on the contrary that S is finite. Since q2 “ 7 ă 13 “ q3 and q3 “ 13 ą 7 “ q4, the set S is non-empty. Since it is finite, we can consider its largest element, say m.Note that it is impossible that qm ą qm`1 ą qm`2 ą . . . because all these numbers are positive integers, so there exists a k ě m such that qk ă qk`1 (rec...	{ "page_id": null, "source": 6828, "title": "from dpo" }
1q2 ` p t ´ 1q ` 1 implies that p \| p p ´ tq2 ` p p ´ tq ` 1, so p and qp´t are prime divisors of pp ´ tq2 ` p p ´ tq ` 1. But p ´ t ą t ´ 1 ě k, so qp´t ą qt´1 “ p and p ¨ qp´t ą p2 ą p p ´ tq2 ` p p ´ tq ` 1, a contradiction. 56 IMO 2013 Colombia ## N4. Determine whether there exists an infinite sequence of nonzero d...	{ "page_id": null, "source": 6828, "title": "from dpo" }
by γ.Now, for each k ě n we have pxk`1 ´ xkqp xk`1 ` xkq “ x2 > k`1 ´ x2 > k “ yk`1 ´ yk “ ak`1 ¨ 10 k. One of the numbers xk`1 ´ xk and xk`1 ` xk is not divisible by 5 γ`1 since otherwise one would have 5γ`1 \| `pxk`1 ´ xkq ` p xk`1 ` xkq˘ “ 2xk`1. On the other hand, we have 5 k \| p xk`1 ´ xkqp xk`1 ` xk q,so 5 k´γ div...	{ "page_id": null, "source": 6828, "title": "from dpo" }
Ak ´ Bk and Ak ` Bk is not divisible by 4. Therefore (1) yields Ak ´ Bk “ 2 and Ak ` Bk “ 22k, hence Ak “ 22k´1 ` 1 and thus x2k`2 “ 5k`1Ak “ 10 k`1 ¨ 2k´2 ` 5k`1 ą 10 k`1, since k ě 2. This implies that y2k`2 ą 10 2k`2 which contradicts the fact that y2k`2 contains 2 k ` 2digits. The desired result follows. Shortliste...	{ "page_id": null, "source": 6828, "title": "from dpo" }
p5kq2 6and hence p5{2qk ă 60 which is false for sufficiently large k. So, assuming that k is large, we get that 2 k´1 divides one of the numbers pk and qk. Hence tpk, q ku “ t 2k´1 ¨ 5rk bk, 2 ¨ 5k´rk cku with nonnegative integers bk , ck, rk such that bkck “ ak`1.Moreover, from (3) we get 6 ą 2k´1 ¨ 5rk bk 2 ¨ 5k´rk c...	{ "page_id": null, "source": 6828, "title": "from dpo" }
2 s ă 2pk ´ rkq ă ` 2 > α ´ 2˘rk ´ 2c1 > α ; if s is large this implies rk ą s, so (6) also holds modulo 10 s. Then (6) and the lemma give yk`1 ” 52pk´rk qc2 > k “ 5s`2s ¨ 5dc2 > k ” 5s ¨ 5dc2 > k pmod 10 sq. (7) By (5) and the minimality of k we have d ď 2C, so 5 dc2 > k ď 52C ¨ 81 “ D. Using 5 4 ă 10 3 we obtain 5s ¨...	{ "page_id": null, "source": 6828, "title": "from dpo" }
ends after at most n steps, and consequently there always has to be some player possessing a winning strategy. So if some n ě k is bad, then Ana has a winning strategy in the game with starting number n.More precisely, if n ě k is such that there is a good integer m with n ą m ě k and gcd pm, n q “ 1, then n itself is ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
Proof. Since rs is bad, there is a move rs ÝÑ x for some good x. Evidently x is coprime to r2s as well, and hence the move r2s ÝÑ x shows that r2s is indeed bad. l Claim 3. If p ą k is prime and n ě k is bad, then np is also bad. Proof. Otherwise we choose a counterexample with n being as small as possible. In particul...	{ "page_id": null, "source": 6828, "title": "from dpo" }
call two integers a, b ě k similar if they are divisible by the same prime numbers not exceeding k. We are to prove that if a and b are similar, then either both of them are good or both are bad. As in this case the product ab is similar to both a and b, it suffices to show the following: if c ě k is similar to some of...	{ "page_id": null, "source": 6828, "title": "from dpo" }
no big prime factors. Proof. Unless b has a big prime factor we may simply choose x “ b. Now let p and q denote a small and a big prime factor of b, respectively. Let a be the product of all small prime factors of b. Further define n to be the least non–negative integer for which the number x “ pna is at least as large...	{ "page_id": null, "source": 6828, "title": "from dpo" }
a and r, contrary to a ÝÑ r being a move. Thereby the problem is solved. > Comment 1. Having reached Claim 4 of Solution 2, there are various other ways to proceed. For instance, one may directly obtain the following fact, which seems to be interesting in its own right: > Claim 5. Any two good numbers have a common sma...	{ "page_id": null, "source": 6828, "title": "from dpo" }
There are infinitely many good numbers, e.g. all multiples of k. The increasing sequence b0, b 1, . . . , of all good numbers may be constructed recursively as follows: ‚ Start with b0 “ k. ‚ If bn has just been defined for some n ě 0, then bn`1 is the smallest number b ą bn that is coprime to none of b0, . . . , b n.T...	{ "page_id": null, "source": 6828, "title": "from dpo" }
show that the word Wk introduced in the previous comment is indeed periodic. The Problem Selection Committee thinks that the above version of the problem is somewhat easier, even though it demands to prove a stronger result. Shortlisted problems – solutions 61 ## N6. Determine all functions f : Q ÝÑ Z satisfying f ˆf p...	{ "page_id": null, "source": 6828, "title": "from dpo" }
absolute value of f pmq ´ m,respectively. Given any integer r, we may plug the triple pm, rb ´ C, b q into (1), thus getting f prq “ f pr ´ ηq. Starting with m and using induction in both directions, we deduce from this that the equation f prq “ C holds for all integers r. Now any rational number y can be written in th...	{ "page_id": null, "source": 6828, "title": "from dpo" }
to be odd and we can substitute ` 1 > 2 , a´1 > 2 , b > 2 ˘ into (1), which yields f ˆω ` p a ´ 1q{ 2 b{2 ˙ “ f ´a b ¯ ‰ ω. (3) 62 IMO 2013 Colombia Recall that 0 ď p a ´ 1q{ 2 ă b{2. Thus, in both cases ω “ 0 and ω “ 1, the left-hand part of (3) equals ω either by the minimality of b, or by f pωq “ ω. A contradiction....	{ "page_id": null, "source": 6828, "title": "from dpo" }
0, then f pxq “ txu holds for all rational x with 0 ď x ă 1 and hence by (2) this even holds for all rational numbers x. Similarly, if ω “ 1, then f pxq “ rxs holds for all x P Q. Thereby the problem is solved. Comment 1. An alternative treatment of Steps B and C from the second case, due to the proposer, proceeds as f...	{ "page_id": null, "source": 6828, "title": "from dpo" }
“ r xs. To complete the argument we just need to observe that if some y P Q satisfies f pyq “ r ys, then (5) yields f ` y > 2 ˘ “ f ´rys > 2 ¯ “ ” rys > 2 ı “ “ y > 2 ‰.Shortlisted problems – solutions 63 Solution 2. Here we just give another argument for the second case of the above solution. Again we use equation (2)...	{ "page_id": null, "source": 6828, "title": "from dpo" }
we get 0 P S Ď p´ 1, `1q and hence the real number α “ sup pSq exists and satisfies 0 ď α ď 1. Let us assume that we actually had 0 ă α ă 1. Note that f pxq “ 0 if x P p 0, α q X Q by (8), and f pxq “ 1 if x P p α, 1q X Q by (9) and (2). Let K denote the unique positive integer satisfying Kα ă 1 ď p K ` 1qα. The first ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
P Q due to (2). Similarly, α “ 1 entails S “ r 0, 1q X Q and f pxq “ txu for all x P Q. Thereby the solution is complete. 64 IMO 2013 Colombia Comment 2. It seems that all solutions to this problems involve some case distinction separating the constant solutions from the unbounded ones, though the “descriptions” of the...	{ "page_id": null, "source": 6828, "title": "from dpo" }
usual, by txu and rxs we denote the integer numbers such that x ´ 1 ă txu ď x and x ď rxs ă x ` 1.) Prove that the number of excellent pairs is equal to the sum of the positive divisors of m. (U.S.A.) Solution. For positive integers a and b, let us denote f pa, b q “ arbν s ´ btaν u. We will deal with various values of...	{ "page_id": null, "source": 6828, "title": "from dpo" }
than f pa, b q by one of a and b. Thus, exactly one of the pairs pa ` b, b q and pa, b ` aq is excellent (for an appropriate value of m). Now let us say that the pairs pa ` b, b q and pa, b ` aq are the children of the pair pa, b q, while this pair is their parent . Next, if a pair pc, d q can be obtained from pa, b q ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
m and min ta, b u ď m. Then all the ancestors of the elements in Sm are again in Sm, and each element in Sm either is of the form pa, a q with a ď m, or has a unique ancestor of this form. From the arguments above we see that all m-excellent pairs lie in Sm.We claim now that the set Sm is finite. Indeed, assume, for in...	{ "page_id": null, "source": 6828, "title": "from dpo" }
m ´ f pa, b q as t “ ka ` ℓb with k and ℓ being some nonnegative integers. Proof. We proceed by induction on the number N of descendants of pa, b q in Sm. If N “ 0 then clearly gpa, b q “ 0. Assume that hpa, b q ą 0; without loss of generality, we have a ď b. Then, clearly, m ´ f pa, b q ě a, so f pa, b ` aq ď f pa, b ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
hpa, b ` aq representations of m ´ f pa, b ` aq “ m ´ f pa, b q as the sum k1a ` ℓ1pb ` aq provides the representation m ´ f pa, b q “ ka ` ℓb with k “ k1 ` ℓ1 ě ℓ1 “ ℓ. This correspondence is obviously bijective, so hpa, b q “ hpa ` b, b q ` hpa, b ` aq “ gpa, b q, as required. Finally, if f pa, b q` b “ m then pa`b, ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
x\|m, x ăm m x “ 1 ` ÿ > d\|m, d ą1 d “ ÿ > d\|m d, as required. Shortlisted problems – solutions 67 Comment. Let us present a sketch of an outline of a different solution. The plan is to check that the number of excellent pairs does not depend on the (irrational) number ν, and to find this number for some appropriate val...	{ "page_id": null, "source": 6828, "title": "from dpo" }
is good, then the point Cν pbq lies on fν paq, which means that the point of fν paq with abscissa b lies between ℓν and ℓ` > ν and is integral. If in addition the pair pa, b ´ aq is not good, then the point of fν paq with abscissa b ´ a lies above ℓ` > ν (see Fig. 1). Thus, the pair pa, b q with b ą a is excellent exac...	{ "page_id": null, "source": 6828, "title": "from dpo" }
(b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b)Cν (b) Pν (a) ℓν ℓ+ > ν fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a)fν (a...	{ "page_id": null, "source": 6828, "title": "from dpo" }
(a)Fν (a)Fν (a)Fν (a)Fν (a)Fν (a)Fν (a)Fν (a)Fν (a)Fν (a) Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (b)Qν (...	{ "page_id": null, "source": 6828, "title": "from dpo" }
number of excellent pairs changes as ν grows. (Having done that, one may find this number for one appropriate value of ν; for instance, it is relatively easy to make this calculation for ν P `1, 1 ` 1 > m ˘.) Consider, for the initial value of ν, some excellent pair pa, t q with a ą t. As ν grows, this pair eventually ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
points of this set, which are exactly the points when the line ℓν passes through some point of the form Cν pbq.The same ideas may be explained in an algebraic language instead of a geometrical one; the same technicalities remain in this way as well.	{ "page_id": null, "source": 6828, "title": "from dpo" }
Title: URL Source: Markdown Content: # MATH 233 - Linear Algebra I Lecture Notes # Cesar O. Aguilar > Department of Mathematics SUNY Geneseo Lecture 0 # Contents 1 Systems of Linear Equations 1 1.1 What is a system of linear equations? . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Matrices . . . . . . . . . . . . ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
. . . . . . . . . . . . 19 3.2 The linear combination problem . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.3 The span of a set of vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4 The Matrix Equation Ax = b 31 4.1 Matrix-vector multiplication . . . . . . . . . . . . . . . . . . . . . . . . . ...	{ "page_id": null, "source": 6828, "title": "from dpo" }
. . . . . . . . . . . . . . . . . . . 49 6.2 The maximum size of a linearly independent set . . . . . . . . . . . . . . . . 53 7 Introduction to Linear Mappings 57 7.1 Vector mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 7.2 Linear mappings . . . . . . . . . . . . . . . . . . . . . . . ....	{ "page_id": null, "source": 6828, "title": "from dpo" }

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