Let $n$ be a positive integer. A number $A$ consists of $2n$ digits, each of which is 4; and a number $B$ consists of $n$ digits, each of which is 8. Prove that $A+2B+4$ is a perfect square.

Click for solution let $x$ be a number with n digits which are all 1's. Then $A=4x(10^n+1)$, $B=8x$ so \begin{eqnarray*} A+2B+4 &=& 4x(10^n+1)+16x+4 \\ \ &=& 4x(10^n-1)+8x +16x+4 \\ \ &=& 4x\cdot 9x +24x +4 \\ \ &=& (6x+2)^2. \end{eqnarray*}

Suppose there are $n$ points in a plane no three of which are collinear with the property that if we label these points as $A_1,A_2,\ldots,A_n$ in any way whatsoever, the broken line $A_1A_2\ldots A_n$ does not intersect itself. Find the maximum value of $n$. Dinu Serbanescu, Romania

Let $D$, $E$, $F$ be the midpoints of the arcs $BC$, $CA$, $AB$ on the circumcircle of a triangle $ABC$ not containing the points $A$, $B$, $C$, respectively. Let the line $DE$ meets $BC$ and $CA$ at $G$ and $H$, and let $M$ be the midpoint of the segment $GH$. Let the line $FD$ meet $BC$ and $AB$ at $K$ and $J$, and let $N$ be the midpoint of the segment $KJ$. a) Find the angles of triangle $DMN$; b) Prove that if $P$ is the point of intersection of the lines $AD$ and $EF$, then the circumcenter of triangle $DMN$ lies on the circumcircle of triangle $PMN$.

Click for solution We will first solve problem b), then problem a). The points D, E, F lie on the angle bisectors of triangle ABC, i. e. on the lines AI, BI, CI, where I is the center of the incircle of triangle ABC. [This is because the angle bisector of an angle in a triangle always passes through the midpoint of the arc cut off from the circumcircle by the opposite side.] Now, we consider the triangle DEF. We have < DIB = 180° - < AIB = < IAB + < IBA (sum of angles in triangle AIB) = A/2 + B/2, while < IDB = < ADB = < ACB (cyclic) = C = 180° - A - B = 180° - 2 (A/2 + B/2) = 180° - 2 < DIB. Thus, < DBI = 180° - < DIB - < IDB = 180° - < DIB - (180° - 2 < DIB) = - < DIB + 2 < DIB = < DIB, so that triangle BDI is isosceles, i. e. we have BD = ID. Similarly, CD = ID. Hence, BD = ID = CD, and the point D is the circumcenter of triangle BIC. Similarly, E and F are the circumcenters of triangles CIA and AIB. [This is an important fact: that the midpoint of the arc BC on the circumcircle of triangle ABC is the circumcenter of triangle BIC.] Now, since the points E and F are the circumcenters of triangles CIA and AIB, both of them lie on the perpendicular bisector of the segment AI. Hence, the line EF is the perpendicular bisector of the segment AI. Similarly, the lines FD and DE are the perpendicular bisectors of the segments BI and CI. In other words, the lines EF, FD, DE pass through the midpoints of the segments AI, BI, CI, respectively, and are perpendicular to these segments. Hence, the point P, being defined as the point of intersection of EF and AI (you have said AD, but this is the same line), must be the midpoint of AI. Now, since DE is perpendicular to CI, the line CI is an altitude of triangle GCH. On the other hand, it is obviously an angle bisector of this triangle. But if an altitude and an angle bisector of a triangle coincide, then the triangle is isosceles; hence, the triangle GCH is isoscles, and the midpoint M of its base GH must lie on the altitude from the apex C. In other words, the point M lies on the line CI. But since the point M also lies on the line DE, and the line DE is the perpendicular bisector of CI, the point M must be the midpoint of CI. Similarly, N is the midpoint of BI. Thus, we have found: The points P, N, M are the midpoints of the segments AI, BI, CI, respectively. Now, since N and M are the midpoints of the sides BI and CI of triangle BIC, we have MN || BC. So we have proven that the line MN is parallel to the line BC. This is the first important assertion. Now we will show that the circumcenter of triangle DMN lies on the circumcircle of triangle PMN. Since the lines EF, FD, DE are perpendicular to the lines AI, BI, CI, respectively, the lines AI, BI, CI are the three altitudes of the triangle DEF. Consequently, the point I is the orthocenter of this triangle, and the points P, N and M are the feet of the altitudes of this triangle. Hence, the circumcircle of triangle PMN is the nine-point circle of triangle DEF. Therefore, it remains to show that the circumcenter of triangle DMN lies on the nine-point circle of triangle DEF. But since < DNI = 90° and < DMI = 90°, the points N and M lie on the circle with diameter DI; hence, the circumenter of triangle DMN is the center of the circle with diameter DI, i. e. the midpoint of the segment DI, hence the midpoint of the segment joining a vertex of triangle DEF with its orthocenter. But it is well-known that such midpoints always lie on the nine-point circle. Hence, the circumenter of triangle DMN lies on the nine-point circle of triangle DEF, i. e. on the circumcircle of triangle PMN. Thus, problem b) is solved. Now, problem a) is simple: Since MN || BC, we have < DMN = < DGK. Now, since the line DE is the perpendicular bisector of the segment CI, and the point M is the midpoint of the segment CI, the triangle CMG is right-angled at M, so we have < CGM = 90° - < GCM. Finally, < GCM = < BCI = C/2. Thus, < DMN = < DGK = < CGM = 90° - < GCM = 90° - C/2. Similarly, < DNM = 90° - B/2. Thus, by the sum of the angles in triangle DMN, we have < MDN = 180° - < DMN - < DNM = 180° - (90° - C/2) - (90° - B/2) = (B + C) / 2 = (180° - A) / 2 (sum of angles in triangle ABC) = 90° - A/2. Hence, we found all three angles of triangle DMN. Problem a) is solved. Darij

Let $x, y, z > -1$. Prove that \[ \frac{1+x^2}{1+y+z^2} + \frac{1+y^2}{1+z+x^2} + \frac{1+z^2}{1+x+y^2} \geq 2. \] Laurentiu Panaitopol