A triangle of sides $\frac{3}{2}, \frac{\sqrt{5}}{2}, \sqrt{2}$ is folded along a variable line perpendicular to the side of $\frac{3}{2}.$ Find the maximum value of the coincident area.

Let $v_0 = 0, v_1 = 1$ and $v_{n+1} = 8 \cdot v_n - v_{n-1},$ $n = 1,2, ...$. Prove that in the sequence $\{v_n\}$ there aren't terms of the form $3^{\alpha} \cdot 5^{\beta}$ with $\alpha, \beta \in \mathbb{N}.$

Click for solution Do you mean $\mathbb{N}=\{1,2,...\}$ ? If yes then the proof is: $3|v_n <=> 3|n$ and $7|v_n <=> 3|n$ - what's easy to check. So if $3|v_n$ then $v_n$ can't be the given form.

Find the greatest $n$ such that $(z+1)^n = z^n + 1$ has all its non-zero roots in the unitary circumference, e.g. $(\alpha+1)^n = \alpha^n + 1, \alpha \neq 0$ implies $|\alpha| = 1.$

Click for solution n equals 7. For $n=7$ it is not difficult to see that all roots have modulus 1 ( use well-known trick to factorize $ (x+y)^7-x^7-y^7$). To prove this one is best, use Gauss theorem: the root of the derivative of a polynomial is in the convex hull of the roots of the polynomial. Thus, if all roots of a polynomial have modulus at most 1, so does the derivative of the polynomial. Just compute the derivative and find it's roots.

Given triangle $ABC$, squares $ABEF, BCGH, CAIJ$ are constructed externally on side $AB, BC, CA$, respectively. Let $AH \cap BJ = P_1$, $BJ \cap CF = Q_1$, $CF \cap AH = R_1$, $AG \cap CE = P_2$, $BI \cap AG = Q_2$, $CE \cap BI = R_2$. Prove that triangle $P_1 Q_1 R_1$ is congruent to triangle $P_2 Q_2 R_2$.

Click for solution The triangle pairs $\triangle ABI \cong \triangle AFC,\ \triangle BCE \cong \triangle BHA,\ \triangle CAG \cong \triangle CJB$ are congruent by SAS, for example, the 1st pair has the sides AB = AF and AI = AC equal, because ABEF and CAIJ are both squares and the angles $\angle BAI = \angle FAC = \angle A + 90^o$ are also equal. Hence, the other corresponding angles of these congruent triangle pairs are also equal. Moreover, the triangle $\triangle ABI$ is obtained by rotating the triangle $\triangle AFC$ around the vertex A by the angle $\angle FAB = 90^o$ and similarly for the other 2 triangle pairs. Hence, the line pairs $IB \perp FC,\ EC \perp AH,\ GA \perp BJ$ are perpendicular to each other. The angles $\angle P_1Q_1R_1 = \angle P_2Q_2R_2$ are equal, because $\angle P_1Q_1R_1 = \widehat{(P_1Q_1, R_1Q_1)} \equiv \widehat{(BJ, FC)} =$ $= \widehat{(GA, FC)} \equiv \widehat{(P_2Q_2, R_2Q_2)} = \angle P_2Q_2R_2$ and similarly, the angles $\angle Q_1R_1P_1 = \angle Q_2R_2P_2,\ \angle R_1P_1Q_1 = \angle R_2P_2Q_2$ are also equal. Thus we established that the triangles $\triangle P_1Q_1R_1 \sim \triangle P_2Q_2R_2$ are similar. Let K, L, M be the diagonal intersections of the squares ABFE, BCGH, CAIJ and X, Y, Z the intersections of the perpendicular line pairs $IB \perp FC,\ EC \perp AH,\ GA \perp BJ$. The quadrilateral AXCI is cyclic, because the angles $\angle AIX \equiv \angle AIB = \angle ACF \equiv \angle ACX$ of the congruent triangles $\triangle ABI \cong \triangle AFC$ are equal. Since the angle $\angle CXI = 90^o$ is right, the circumcenter of this cyclic quadrilateral is the midpoint of the chord CI, the diagonal of the square CAIJ. Hence, the circumcircle of the quadrilateral AXCI is identical with the circumcircle $(M)$ of the square CAIJ and the square vertex J also lies on this circle. Since AJ is a diameter of the circle $(K)$ and the angle $\angle AZJ = 90^o$ is also right, the point Z also lies of this circle. The power of the triangle vertex B to the circle $(K)$ is equal to $BX \cdot BI = BZ \cdot BJ$, which implies $\frac{BI}{BJ} = \frac{BZ}{BX}$ The right angle triangles $\triangle BXQ_1 \sim \triangle BZQ_2$, because the angles $\angle BXQ_1 = \angle BZQ_2 = 90^o$ are both right and the angles $\angle BQ_1X \equiv \angle P_1Q_1R_1 = \angle P_2Q_2R_2 \equiv \angle ZQ_2B$ of similar triangles $\triangle P_1Q_1R_1 \sim \triangle P_2Q_2R_2$ are equal. Consequently, $\frac{BQ_2}{BQ_1} = \frac{BZ}{BX} = \frac{BI}{BJ}$ But this means that the triangles $\triangle BQ_2Q_1 \sim \triangle BIJ$ are centrally similar with the similarity center B and thus the lines $Q_1Q_2 \parallel JI \parallel CA$ are parallel. In a similar way, we could show that the lines $P_1P_2 \parallel HG \parallel BC$ and the lines $R_1R_2 \parallel FE \parallel AB$ are also parallel. Let S be the concurrency point of the lines AL, BM, CK, lying (together with the centroid and the 1st Fermat point of the triangle $\triangle ABC$) on Kiepert's hyperbola. Since the angles $\angle AMC = \angle AYC = 90^o$ are both right, the quadrilateral AYCM is cyclic. Since N is the midpoint of the are AC of its circumcircle opposite to the vertex Y, the line YC bisects the right angle $\angle AYC$. Let $(M_1), (M_2)$ be 2 centrally similar circles on centrally similar diameters $IJ \sim Q_1Q_2$ with the similarity center B. Since the right angle triangle $\triangle IMJ$ with the circumcircle $(M_1)$ is isosceles with IM = JM, the line BM cuts the corresponding arcs $JI, Q_1Q_2$ of the circles $(M_1), (M_2)$ at their midpoints M, S', the right angle triangle $\triangle Q_1S'Q_2$ is also isosceles with $Q_1S' = Q_2S'$ and the angles $\angle Q_2Q_1S = \angle Q_1Q_2S' = 45^o$. Since the angles $\angle Q_1XQ_2 = \angle Q_1ZQ_2 = 90^o$ formed by the perpendicular lines $Q_1X \equiv FC \perp IB \equiv Q_2X$ and $Q_1Z \equiv JB \perp AG \equiv Q_2Z$ are both right, the points $X, Z$ both lie on the circle $(M_2)$. Since the angle $\angle AZK \equiv \angle Q_2ZS = 45^o$, this angle spans the same arc $Q_2S'$ as the angle $\angle Q_2Q_1S' = 45^o$ and as a result, the points $S' \equiv S$ are identical. Thus we proved that the triangle $\triangle Q_1SQ_2$ is isosceles with the angle $\angle Q_1SQ_2 = 90^o$ and $Q_1S = Q_2S$. In a similar way, we could prove that the triangles $\triangle P_1SP_2,\ \triangle R_1SR_2$ are also isosceles with the angles $\angle P_1SP_2 = \angle R_1SR_2 = 90^o$ and $P_1S = P_2S,\ R_1S = R_2S$. As a result, the triangle $\triangle P_2Q_2R_2$ is obtained from the triangle $\triangle P_1Q_1R_1$ by a rotation around the center S by the angle $\angle P_1SP_2 = \angle Q_1SQ_2 = \angle R_1SR_2 = 90^o$, which means that these 2 triangles are congruent. Yetti

Let $\mathbb{N} = \{1,2, \ldots\}.$ Does there exists a function $f: \mathbb{N} \mapsto \mathbb{N}$ such that $\forall n \in \mathbb{N},$ $f^{1989}(n) = 2 \cdot n$ ?

Click for solution Let $n_1,n_2,\ldots$ be the increasing sequence of odd numbers $\ge 1$, put, for all $k\ge 0$, $f(n_{1989k+i})=n_{1989k+i+1},\ \forall i\in\overline{1,1988}$, $f(n_{1989k+1989})=2n_{1989k+1}$, and then extend the function to even numbers by using $f(2n)=2f(n),\ \forall n\ge 1$.

$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$. Alternative formulation. Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

Click for solution My solution uses another approach: orl wrote: $AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC + AD - AB - AC = 0$, find the range of $\angle BAC$. First we show: Lemma 1. Let D be the foot of the altitude to the side BC of a triangle ABC. Then, BC + AD - AB - AC = 0 holds if and only if $\tan\frac{B}{2}+\tan\frac{C}{2}=1$. Proof of Lemma 1. Let $s=\frac{a+b+c}{2}$ be the semiperimeter, $\Delta$ the area and $r_a$ the A-exradius of triangle ABC. Then, on one hand, $\Delta=\frac12\cdot a\cdot AD$, since AD is the altitude to the side BC of triangle ABC, but on the other hand, $\Delta=\left(s-a\right)r_a$ by a well-known formula. Hence, $\frac12\cdot a\cdot AD=\left(s-a\right)r_a$; equivalently, $AD=\frac{2\left(s-a\right)r_a}{a}$. On the other hand, b + c - a = (a + b + c) - 2a = 2s - 2a = 2 (s - a). Thus, $BC+AD-AB-AC=BC+\frac{2\left(s-a\right)r_a}{a}-AB-AC=a+\frac{2\left(s-a\right)r_a}{a}-c-b$ $=\frac{2\left(s-a\right)r_a}{a}-\left(b+c-a\right)=\frac{2\left(s-a\right)r_a}{a}-2\left(s-a\right)$ $=2\left(s-a\right)\frac{r_a-a}{a}$. Since s - a > 0, we thus have BC + AD - AB - AC = 0 if and only if $r_a-a=0$, what is equivalent to $a=r_a$. Now, let the A-excircle of triangle ABC have the center $I_a$ and touch the side BC at a point X. Then, $I_aX=r_a$, and $I_aX\perp BC$. But, being the A-excenter of triangle ABC, the point $I_a$ lies on the external angle bisector of the angle ABC, and thus $\measuredangle I_aBX=90^{\circ}-\frac{B}{2}$. Hence, in the right-angled triangle $BXI_a$, we have $BX=I_aX\cdot\cot\measuredangle I_aBX=r_a\cdot\cot\left(90^{\circ}-\frac{B}{2}\right)=r_a\cdot\tan\frac{B}{2}$. Similarly, $CX=r_a\cdot\tan\frac{C}{2}$. Thus, $a=BC=BX+CX=r_a\cdot\tan\frac{B}{2}+r_a\cdot\tan\frac{C}{2}=r_a\cdot\left(\tan\frac{B}{2}+\tan\frac{C}{2}\right)$. Hence, $a=r_a$ is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$. But we saw before that BC + AD - AB - AC = 0 is equivalent to $a=r_a$. Hence, BC + AD - AB - AC = 0 is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$, and Lemma 1 is proven. Another lemma now: Lemma 2. For a given angle x such that 0° < x < 90°, the existance of two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$ is equivalent to $45^{\circ}<x\leq 2\arctan\frac12$. Note. $2\arctan\frac12\approx 53,13^{\circ}$. Proof of Lemma 2. First, we show that if there exist two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$, then $45^{\circ}<x\leq 2\arctan\frac12$. In fact, the function f(t) = tan t is convex on the interval [0°; 90°[; hence, $\tan y+\tan z\geq 2\tan\frac{y+z}{2}$. In other words, $1\geq 2\tan\frac{x}{2}$. Hence, $\frac12\geq\tan\frac{x}{2}$, so that $\arctan\frac12\geq\frac{x}{2}$ and $2\arctan\frac12\geq x$. On the other hand, by the tangens addition formula, $\tan\left(y+z\right)=\frac{\tan y+\tan z}{1-\tan y\tan z}$. In other words, $\tan x=\frac{1}{1-\tan y\tan z}$. Since tan y and tan z are positive (as 0° < y < 90° and 0° < z < 90°), we have 1 - tan y tan z < 1; but tan x is also positive (since 0° < x < 90°), and thus, tan x > 1, and thus x > 45°. Combining this with $2\arctan\frac12\geq x$, we obtain $45^{\circ}<x\leq 2\arctan\frac12$. Remains to prove the converse: If $45^{\circ}<x\leq 2\arctan\frac12$, then there exist two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$. In fact, first define the angles y and z as follows: y = 0°, z = x. This will be called the "starting position". Of course, y + z = x is satisfied in this position, but $\tan y+\tan z=1$ does not hold. Now, let's start continuously increasing y and decreasing z at the same speed, so that the sum y + z = x remains invariant. Then, the angles y and z come closer and closer to each other, and at the end both of them come together: $y=z=\frac{x}{2}$. This will be called the "ending position". Now look at what happens with the value of tan y + tan z while we increase y and decrease z. Since the function f(t) = tan t is continous on the interval [0°; 90°[, this value of tan y + tan z behaves continuously. In the "starting position", the value of tan y + tan z equals tan 0° + tan x = tan x; this is > 1, since x > 45°. In the "ending position", the value of tan y + tan z equals $\tan\frac{x}{2}+\tan\frac{x}{2}=2\tan\frac{x}{2}\leq2\tan\arctan\frac12$ (since $x\leq 2\arctan\frac12$ yields $\frac{x}{2}\leq\arctan\frac12$) $=2\cdot\frac12=1$. Hence, by the intermediate value theorem, somewhere between the "starting position" and the "ending position", the expression tan y + tan z must take the value 1. Thus, we have proved the existence of two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$. Hence, Lemma 2 is proven. Now, the problem becomes immediate: After Lemma 1, the condition BC + AD - AB - AC = 0 is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$. Hence, we are looking for the range of angles A such that 0° < A < 180° and such that there exist angles B and C with 0° < B < 180°, 0° < C < 180° and A + B + C = 180° (these are the conditions for the existence of a triangle with angles A, B, C) which satisfy $\tan\frac{B}{2}+\tan\frac{C}{2}=1$. Set $x=90^{\circ}-\frac{A}{2}$, $y=\frac{B}{2}$, $z=\frac{C}{2}$; then, the conditions 0° < A < 180°, 0° < B < 180°, 0° < C < 180° rewrite as 0° < x < 90°, 0° < y < 90°, 0° < z < 90°, the condition A + B + C = 180° rewrites as y + z = x (since $y+z-x=\frac{B}{2}+\frac{C}{2}-\left(90^{\circ}-\frac{A}{2}\right)=\frac{A+B+C}{2}-90^{\circ}=\frac{A+B+C-180^{\circ}}{2}$), and the condition $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ rewrites as $\tan y+\tan z=1$. Now, by Lemma 2, the existence of two angles y and z satisfying these conditions is equivalent to $45^{\circ}<x\leq 2\arctan\frac12$. Since $x=90^{\circ}-\frac{A}{2}$, this rewrites as $45^{\circ}<90^{\circ}-\frac{A}{2}\leq 2\arctan\frac12$. Upon subtraction from 90°, this becomes $45^{\circ}>\frac{A}{2}\geq 90^{\circ}-2\arctan\frac12$, and multiplication with 2 transforms this into $90^{\circ}>A\geq 2\left(90^{\circ}-2\arctan\frac12\right)$. This rewrites as $90^{\circ}>A\geq 180^{\circ}-4\arctan\frac12$. Thus, the range of our angle A is $\left[180^{\circ}-4\arctan\frac12;\;90^{\circ}\right[$. Note that $180^{\circ}-4\arctan\frac12\approx 73,74^{\circ}$. Darij

$1989$ equal circles are arbitrarily placed on the table without overlap. What is the least number of colors are needed such that all the circles can be painted with any two tangential circles colored differently.

$\forall n \in \mathbb{N}$, $P(n)$ denotes the number of the partition of $n$ as the sum of positive integers (disregarding the order of the parts), e.g. since $4 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2 = 4$, so $P(4)=5$. "Dispersion" of a partition denotes the number of different parts in that partitation. And denote $q(n)$ is the sum of all the dispersions, e.g. $q(4)=1+2+2+1+1=7$. $n \geq 1$. Prove that (1) $q(n) = 1 + \sum^{n-1}_{i=1} P(i).$ (2) $1 + \sum^{n-1}_{i=1} P(i) \leq \sqrt{2} \cdot n \cdot P(n)$.

Click for solution the first part is rather trivial: $q(n)=\sum_\lambda \text{disp}(\lambda)$ as $\lambda$ runs over all the partitions of $n$ and $\text{disp}(\lambda)$ counts the number of distinct parts in $\lambda$. clearly this is the same as counting for each $i$, the number of partitions of $n$ that count a distinct part of size $i$.If $i=n$ then of course there is only one such partition. else, the number of partitions that have a part of size $i\ = p(n-i)$, so $q(n)=1+\sum_{i=1}^{n-1} p(n-i)$ and that is what we need. the second part is easily improved by the fact that $np(n)=\sum_{i=1}^n \sigma(i)p(n-i)$ where $\sigma(i)$ is the sum of all positive divisors of $i$.