1、The 9th Romanian Master of Mathematics CompetitionDay 2 | SolutionsProblem 4. In the Cartesian plane, let G1 and G2 be the graphs of the quadratic functionsf1(x) = p1x2 +q1x+r1 and f2(x) = p2x2 +q2x+r2, where p1 0 p2. The graphs G1 and G2cross at distinct points A and B. The four tangents to G1 and
2、G2 at A and B form a convexquadrilateral which has an inscribed circle. Prove that the graphs G1 and G2 have the same axisof symmetry.Alexey Zaslavsky, RussiaSolution 1. LetAi andBi be the tangents toGi at A and B, respectively, and let Ci =AiBi.Since f1(x) is convex and f2(x) is concave, the convex
3、 quadrangle formed by the four tangentsis exactly AC1BC2.Lemma. If CA and CB are the tangents drawn from a point C to the graph G of a quadratictrinomial f(x) = px2 +qx+r, A;B2G, A6= B, then the abscissa of C is the arithmetic meanof the abscissae of A and B.Proof. Assume, without loss of generality
4、, that C is at the origin, so the equations of the twotangents have the form y = kax and y = kbx. Next, the abscissae xA and xB of the tangencypoints A and B, respectively, are multiple roots of the polynomials f(x) kax and f(x) kbx,respectively. By the Vieta theorem, x2A = r=p = x2B, so xA = xB, si
5、nce the case xA = xB isruled out by A6= B.ABC1C2A A0BC1C2OA BC1C2OThe Lemma shows that the line C1C2 is parallel to the y-axis and the points A and B areequidistant from this line.Suppose, if possible, that the incentre O of the quadrangle AC1BC2 does not lie on theline C1C2. Assume, without loss of
6、 generality, that O lies inside the triangle AC1C2 and let A0be the re ection of A in the line C1C2. Then the ray CiB emanating from Ci lies inside theangle ACiA0, so B lies inside the quadrangle AC1A0C2, whence A and B are not equidistantfrom C1C2 | a contradiction.Thus O lies on C1C2, so the lines
7、 ACi and BCi are re ections of one another in the line C1C2,and B = A0. Hence yA = yB, and since fi(x) = yA + pi(x xA)(x xB), the line C1C2 is theaxis of symmetry of both parabolas, as required.Solution 2. Use the standard equation of a tangent to a smooth curve in the plane, to deducethat the tange
8、nts at two distinct points A and B on the parabola of equation y = px2 +qx+r,1p6= 0, meet at some point C whose coordinates arexC = 12(xA +xB) and yC = pxAxB +q 12(xA +xB) +r:Usage of the standard formula for Euclidean distance yieldsCA = 12jxB xAjp1 + (2pxA +q)2 and CB = 12jxB xAjp1 + (2pxB +q)2;so
9、, after obvious manipulations,CB CA = 2p(xB xA)jxB xAj(p(xA +xB) +q)p1 + (2pxA +q)2 +p1 + (2pxB +q)2:Now, write the condition in the statement in the form C1B C1A = C2B C2A, apply theabove formula and clear common factors to getp1(p1(xA +xB) +q1)p1 + (2p1xA +q1)2 +p1 + (2p1xB +q1)2 =p2(p2(xA +xB) +q
10、2)p1 + (2p2xA +q2)2 +p1 + (2p2xB +q2)2:Next, use the fact that xA and xB are the solutions of the quadratic equation (p1 p2)x2 +(q1 q2)x+r1 r2 = 0, so xA +xB = (q1 q2)=(p1 p2), to obtainp1(p1q2 p2q1)p1 + (2p1xA +q1)2 +p1 + (2p1xB +q1)2 =p2(p1q2 p2q1)p1 + (2p2xA +q2)2 +p1 + (2p2xB +q2)2:Finally, sinc
11、e p1p2 n, in which case the sticks in S span all n columns,and notice that we are again done if jSj n, to assume further jSjn.Let S0 = GhrS, let T be set of all neighbours of S, and let T0 = Gv rT. Since the sticks inS span all n columns, jTj n, so jT0j n 2. Transposition of the above argument (repl
12、ace Sby T0), shows that jT0j jS0j, so jSj jTj.Remark. Here is an alternative argument for s =jSjn. Add to S two empty sticks formallypresent to the left of the leftmost hole and to the right of the rightmost one. Then there are at4least s n+ 2 rows containing two sticks from S, so two of them are se
13、parated by at least s nother rows. Each hole in those s n rows separates two vertical sticks from Gv both of whichare neighbours of S. Thus the vertices of S have at least n+ (s n) neighbours.Solution 4. Yet another proof of the estimate m(A) 2n 2. We use the induction on n. Nowwe need the base case
14、s n = 2;3 which can be completed by hands.Assume now that n 3 and consider any dissection of A into sticks. De ne the cross of ahole as in Solution 1, and notice that each stick is contained in some cross. Thus, if the dissectioncontains more than n sticks, then there exists a cross containing at le
15、ast two sticks. In this case,remove this cross from the sieve to obtain an (n 1) (n 1) sieve. The dissection of the originalsieve induces a dissection of the new array: even if a stick is partitioned into two by the removedcross, then the remaining two parts form a stick in the new array. After this
16、 operation has beenperformed, the number of sticks decreases by at least 2, and by the induction hypothesis thenumber of sticks in the new dissection is at least 2n 4. Hence, the initial dissection contains atleast (2n 4) + 2 = 2n 2 sticks, as required.It remains to rule out the case when the dissec
17、tion contains at most n sticks. This can be donein many ways, one of which is removal a cross containing some stick. The resulting dissection ofan (n 1) (n 1) array contains at most n 1 sticks, which is impossible by the inductionhypothesis since n 1 OQ. Let the line through O and parallel to PQmeet
18、 AB at M, and CB at N. Since OP OQ, the angle SPQ is acute and the angle PQR isobtuse, so the angle AOB is obtuse, the angle BOC is acute, M lies on the segment AB, andN lies on the extension of the segment BC beyond C. Therefore: OA OM, since the angleOMA is obtuse; OM ON, since OM : ON = KP : KQ,
19、where K is the projection of O ontoPQ; and ON OC, since the angle OCN is obtuse. Consequently, OAOC.Similarly, OROS yields OC OA: a contradiction. Consequently, OP = OQ and PQRSis a rectangle. This ends the proof.Solution 2. (Ilya Bogdanov) To begin, we establish a useful lemma.Lemma 2. If P is a po
20、int on the side AB of a triangle OAB, thensinAOPOB +sinPOBOA =sinAOBOP :Proof. Let XYZ denote the area of a triangle XYZ, to write0 = 2(AOB POB POC) = OA OB sinAOB OB OP sinPOB OP OA sinAOP;and divide by OA OB OP to get the required identity.A similar statement remains valid if the point C lies on t
21、he line AB; the proof is obtained byusing signed areas and directed lengths.We now turn to the solution. We rst prove some sort of a converse statement, namely:Claim. Let PQRS be a cyclic quadrangle with O = PRQS; assume that no its diagonal isperpendicular to a side. Let A, B, C, and D be the lines
22、 through O perpendicular to SP,PQ, QR, and RS, respectively. Choose any point A2A and successively de ne B = APB,C = BQC, D = CRD, and A0 = DSA. Then A0 = A.Proof. We restrict ourselves to the case when the points A, B, C, D, and A0 lie on A, B,C, D, and A on the same side of O as their points of in
23、tersection with the respective sides ofthe quadrilateral PQRS. Again, a general case is obtained by suitable consideration of directedlengths.7=2=2 =2 =2 =2 A = A0BCDPQRS OABCDDenote= QPR = QSR = =2 POB = =2 DOS;= RPS = RQS = =2 AOP = =2 QOC;= SQP = SRP = =2 BOQ = =2 ROD;= PRQ = PSQ = =2 COR = =2 SO
24、A:By Lemma 2 applied to the lines APB, PQC, CRD, and DSA0, we getsin( + )OP =cos OA +cos OB ;sin( + )OQ =cos OB +cos OC ;sin( + )OR =cos OC +cos OD;sin( + )OS =cos OD +cos OA0:Adding the two equalities on the left and subtracting the two on the right, we see that therequired equality A = A0 (i.e., c
25、os =OA = cos =OA0, in view of cos 6= 0) is equivalent to therelation sinQPSOP +sinSRQOR =sinPQROQ +sinRSPOS :Let d denote the circumdiameter of PQRS, so sinQPS = sinSRQ = QS=d and sinRSP =sinPQR = PR=d. Thus the required relation readsQSOP +QSOR =PROS +PROQ; orQS PROP OR =PR QSOS OQ:The last relatio
26、n is trivial, due again to cyclicity.Finally, it remains to derive the problem statement from our Claim. Assume that PQRS isnot cyclic, e.g., that OP OR OQ OS, where O = PRQS. Mark the point S0 on the rayOS so that OP OR = OQ OS0. Notice that no diagonal of PQRS is perpendicular to a side,so the qua
27、drangle PQRS0 satis es the conditions of the claim.Let 0A and 0D be the lines through O perpendicular to PS0 and RS0, respectively. Then0A and 0D cross the segments AP and RD, respectively, at some points A0 and D0. By theClaim, the line A0D0 passes through S0. This is impossible, because the segmen
28、t A0D0 crossesthe segment OS at some interior point, while S0 lies on the extension of this segment. Thiscontradiction completes the proof.Remark. According to the author, there is a remarkable corollary that is worth mentioning:Four lines dissect a convex quadrangle into nine smaller quadrangles to make it into a 3 3 array8ABCDPQRS OA0D0S0A BCD0A0D000000of quadrangular cells. Label these cells 1 through 9 from left to right and from top to bottom.If the rst eight cells are orthodiagonal, then so is the ninth.9
