MathDB
Find the number of switched remained on

Source: 2019 Jozsef Wildt International Math Competition-W. 28

5/18/2020
In a room, we have 2019 aligned switches, connected to 2019 light bulbs, all initially switched on. Then, 2019 people enter the room one by one, performing the operation: The first, uses all the switches; the second, every second switch; the third, every third switch, and so on. How many lightbulbs remain switched on, after all the people entered ?
puzzlecombinatorics
Prove this expression

Source: 2019 Jozsef Wildt International Math Competition-W. 29

5/18/2020
Prove that ∫0∞e3t4e4t(3t−1)+2e2t(15t−17)+18(1−t)(1+e4t−e2t)2=12∑k=0∞(−1)k(2k+1)2−10\int \limits_0^{\infty} e^{3t}\frac{4e^{4t}(3t - 1) + 2e^{2t}(15t - 17) + 18(1 - t)}{\left(1 + e^{4t} - e^{2t}\right)^2}=12\sum \limits_{k=0}^{\infty}\frac{(-1)^k}{(2k + 1)^2}-10
integrationSummation
Prove the limit and find the value on zeta function

Source: 2019 Jozsef Wildt International Math Competition-W. 30

5/18/2020
[*] Prove that lim⁡n→∞(n+14−ζ(3)−ζ(5)−⋯−ζ(2n+1))=0\lim \limits_{n \to \infty} \left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots -\zeta(2n+1)\right)=0 [*] Calculate ∑n=1∞(n+14−ζ(3)−ζ(5)−⋯−ζ(2n+1))\sum \limits_{n=1}^{\infty} \left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots -\zeta(2n+1)\right)
zeta functionlimitSummationfunction
Prove this integral inequality

Source: 2019 Jozsef Wildt International Math Competition-W. 31

5/19/2020
Let a,b∈Γa, b \in \Gamma, a<ba < b and the differentiable function f:[a,b]→Γf : [a, b] \to \Gamma, such that f(a)=af (a) = a and f(b)=bf (b) = b. Prove that ∫ab(f′(x))2dx≥b−a\int \limits_{a}^{b} \left(f'(x)\right)^2dx \geq b-a
integrationinequalitiescalculus
Prove this inequality

Source: 2019 Jozsef Wildt International Math Competition-W. 34

5/19/2020
Let aa, bb, cc be positive real numbers and let mm, nn (m≥n)(m \geq n) be positive integers. Prove thatan−1bn−1cm−n−1am+n+bm+n+anbncm−n+bn−1bcn−1am−n−1bm+n+cm+n+bncnam−n+cn−1an−1bm−n−1cm+n+am+n+cnanbm−n≤1abc\frac{a^{n-1}b^{n-1}c^{m-n-1}}{a^{m+n}+b^{m+n}+a^nb^nc^{m-n}}+\frac{b^{n-1}bc^{n-1}a^{m-n-1}}{b^{m+n}+c^{m+n}+b^nc^na^{m-n}}+\frac{c^{n-1}a^{n-1}b^{m-n-1}}{c^{m+n}+a^{m+n}+c^na^nb^{m-n}}\leq \frac{1}{abc}
inequalities
Prove this inequality made by 4 sequences

Source: 2019 Jozsef Wildt International Math Competition-W. 33

5/19/2020
Let 0<1q≤1p<10 < \frac{1}{q} \leq \frac{1}{p} < 1 and 1p+1q=1\frac{1}{p}+\frac{1}{q}=1. Let uku_k, vkv_k, aka_k and bkb_k be non-negative real sequences such as uk2>akpu^2_k > a^p_k and vk>bkqv_k > b^q_k, where k=1,2,⋯ ,nk = 1, 2,\cdots , n. If 0<m1≤uk≤M10 < m_1\leq u_k \leq M_1 and 0<m2≤vk≤M20 < m_2 \leq v_k \leq M_2 , then (∑k=1n(lp(uk+vk)2−(ak+bk)p))1p≥(∑k=1n(uk2−akp))1p(∑k=1n(vk2−bkp))1p\left(\sum \limits_{k=1}^n\left(l^p\left(u_k+v_k\right)^2-\left(a_k+b_k\right)^p\right)\right)^{\frac{1}{p}}\geq \left(\sum \limits_{k=1}^n\left(u_k^2-a_k^p\right)\right)^{\frac{1}{p}}\left(\sum \limits_{k=1}^n\left(v_k^2-b_k^p\right)\right)^{\frac{1}{p}}where l=M1M2+m1m22m1M1m2M2l=\frac{M_1M_2+m_1m_2}{2\sqrt{m_1M_1m_2M_2}}
inequalitiesSummationSequences
Prove this inequality based on 4 squences

Source: 2019 Jozsef Wildt International Math Competition-W. 32

5/19/2020
Let uku_k, vkv_k, aka_k and bkb_k be non-negative real sequences such as uk>aku_k > a_k and vk>bkv_k > b_k, where k=1,2,⋯ ,nk = 1, 2,\cdots , n. If 0<m1≤uk≤M10 < m_1 \leq u_k \leq M_1 and 0<m2≤vk≤M20 < m_2 \leq v_k \leq M_2, then ∑k=1n(lukvk−akbk)≥(∑k=1n(uk2−ak2))12(∑k=1n(vk2−bk2))12\sum \limits_{k=1}^n(lu_kv_k-a_kb_k)\geq \left(\sum \limits_{k=1}^n\left(u_k^2-a_k^2\right)\right)^\frac{1}{2}\left(\sum \limits_{k=1}^n\left(v_k^2-b_k^2\right)\right)^\frac{1}{2}wherel=M1M2+m1m22m1M1m2M2l=\frac{M_1M_2+m_1m_2}{2\sqrt{m_1M_1m_2M_2}}
SummationSequencesinequalities
Prove this cyclic inequality

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
Let aa, bb, c∈(0,+∞)c \in (0,+\infty) . Then the following inequality is true:(a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)+a+b+c≤(ab+bc+ca)(1ab+1bc+1ca)\sqrt{(a+b)(b+c)}+\sqrt{(b+c)(c+a)}+\sqrt{(c+a)(a+b)}+a+b+c\leq \left(ab+bc+ca\right)\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)
inequalitiescyclic inequality
Find the limit of this expression

Source: 2019 Jozsef Wildt International Math Competition-W. 37

5/19/2020
For real a>1a > 1 findlim⁡n→∞∏k=2n(a−a1k)n\lim \limits_{n \to \infty}\sqrt[n]{\prod \limits_{k=2}^n \left(a-a^{\frac{1}{k}}\right)}
limitProduct
Prove this inequalitiy 2

Source: 2019 Jozsef Wildt International Math Competition-W. 36

5/19/2020
For any aa, bb, c>0c > 0 and for any n∈N∗n \in \mathbb{N}^*, prove the inequality(a−b)(ab)n+(b−c)(bc)n+(c−a)(ca)n≥(a−b)ab+(b−c)bc+(c−a)ca(a - b)\left(\frac{a}{b}\right)^n+(b - c)\left(\frac{b}{c}\right)^n+(c - a)\left(\frac{c}{a}\right)^n\geq (a - b)\frac{a}{b}+(b - c)\frac{b}{c}+(c - a)\frac{c}{a}
inequalities
Prove this inequality on area of the triangle

Source: 2019 Jozsef Wildt International Math Competition-W. 38

5/19/2020
Let aa, bb, cc be the sides of an acute triangle △ABC\triangle ABC , then for any x,y,z≥0x, y, z \geq 0, such that xy+yz+zx=1xy+yz+zx=1 holds inequality:a2x+b2y+c2z≥4Fa^2x + b^2y + c^2z \geq 4F where FF is the area of the triangle △ABC\triangle ABC
inequalitiesTriangle
Prove these 3 properties

Source: 2019 Jozsef Wildt International Math Competition-W. 39

5/19/2020
Let uu, vv, ww complex numbers such that: u+v+w=1u + v + w = 1, u2+v2+w2=3u^2 + v^2 + w^2 = 3, uvw=1uvw = 1. Prove that
[*] uu, vv, ww are distinct numbers two by two [*] If S(k)=uk+vk+wkS(k)= u^k + v^k + w^k, then S(k)S(k) is an odd natural number [*] The expressionu2n+1−v2n+1u−v+v2n+1−w2n+1v−w+w2n+1−u2n+1w−u\frac{u^{2n+1} - v^{2n+1}}{u-v}+\frac{v^{2n+1}-w^{2n+1}}{v-w}+\frac{w^{2n+1}-u^{2n+1}}{w-u}is an integer number.
complex numbersnumber theory
Prove this expression on fibonacci numbers is divisible by 150 for any n

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
Let fnf_n be nnth Fibonacci number defined by recurrence fn+1−fn−fn−1=0f_{n+1} - f_n - f_{n-1} = 0, n∈Nn \in \mathbb{N} and initial conditions f0=0f_0 = 0, f1=1f_1 = 1. Prove that for any n∈Nn \in \mathbb{N} (n−1)(n+1)(2nfn+1−(n+6)fn)(n - 1) (n + 1) (2nf_{n+1} - (n + 6) f_n)is divisible by 150 for any n∈Nn \in \mathbb{N}.
Fibonacci sequencenumber theory
Show that number of points inside the tetrahedron

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
For n∈Nn \in \mathbb{N}, consider in R3\mathbb{R}^3 the regular tetrahedron with vertices O(0,0,0)O(0, 0, 0), A(n,9n,4n)A(n, 9n, 4n), B(9n,4n,n)B(9n, 4n, n) and C(4n,n,9n)C(4n, n, 9n). Show that the number NN of points (x,y,z)(x, y, z), [x,y,z∈Z][x, y, z \in \mathbb{Z}] inside or on the boundary of the tetrahedron OABCOABC is given byN=343n33+35n22+7n6+1N=\frac{343n^3}{3}+\frac{35n^2}{2}+\frac{7n}{6}+1
3D geometrycoordinate geometrygeometrytetrahedron
Prove this property

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
For pp, qq, ll strictly positive real numbers, consider the following problem: for y≥0y \geq 0 fixed, determine the values x≥0x \geq 0 such that xp−lxq≤yx^p - lx^q \leq y. Denote by S(y)S(y) the set of solutions of this problem. Prove that if one has p<qp < q, ϵ∈(0,l1p−q)\epsilon \in (0, l^\frac{1}{p-q}), 0≤x≤ϵ0 \leq x \leq \epsilon and x∈S(y)x \in S(y), then x≤kyδ, where k=ϵ(ϵp−lϵq)−1p and δ=1px\leq ky^{\delta},\ \text{where}\ k=\epsilon\left(\epsilon^p-l\epsilon^q\right)^{-\frac{1}{p}}\ \text{and}\ \delta=\frac{1}{p}
set theorynumber theory
Prove this trigonometric inequality with complex numbers

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
Consider the complex numbers a1,a2,⋯ ,ana_1, a_2,\cdots , a_n, n≥2n \geq 2. Which have the following properties:
[*] ∣ai∣=1|a_i|=1 ∀\forall i=1,2,⋯ ,ni=1,2,\cdots , n [*] ∑k=1narg(ak)≤π\sum \limits_{k=1}^n arg(a_k)\leq \pi
Show that the inequality(n2cot⁡(π2n))−1∣∑k=0n(−1)k[3n2−(8k+5)n+4k(k+1)σk]∣≥(1+1n)2cot⁡2(π2n)+16∣∑k=0n(−1)kσk∣\left(n^2\cot \left(\frac{\pi}{2n}\right)\right)^{-1}\left |\sum \limits_{k=0}^n(-1)^k\left[3n^2-(8k+5)n+4k(k+1)\sigma_k\right]\right |\geq \sqrt{\left(1+\frac{1}{n}\right)^2\cot^2 \left(\frac{\pi}{2n}\right)}+16\left |\sum \limits_{k=0}^n(-1)^k\sigma_k\right |where σ0=1\sigma_0=1, σk=∑1≤i1≤i2≤⋯≤ik≤nai1ai2⋯aik\sigma_k=\sum \limits_{1\leq i_1\leq i_2\leq \cdots \leq i_k\leq n}a_{i_1}a_{i_2}\cdots a_{i_k}, ∀\forall k=1,2,⋯ ,nk=1,2,\cdots , n
inequalitiestrigonometrycomplex numbers
Prove for the expression based on limit of sequence of polynomial

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
Consider the sequence of polynomials P0(x)=2P_0(x) = 2, P1(x)=xP_1(x) = x and Pn(x)=xPn−1(x)−Pn−2(x)P_n(x) = xP_{n-1}(x) - P_{n-2}(x) for n≥2n \geq 2. Let xnx_n be the greatest zero of PnP_n in the the interval ∣x∣≤2|x| \leq 2. Show that lim⁡n→∞n2(4−2π+n2∫xn2Pn(x)dx)=2π−4−π312\lim \limits_{n \to \infty}n^2\left(4-2\pi +n^2\int \limits_{x_n}^2P_n(x)dx\right)=2\pi - 4-\frac{\pi^3}{12}
limitintegrationSequencespolynomialalgebra
Prove this property of the matrix

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
We consider a natural number nn, n≥2n \geq 2 and the matrices
\begin{tabular}{cc} A=(123⋯nn12⋯n−1n−1n1⋯n−2⋯⋯⋯⋯⋯234⋯1)A= \begin{pmatrix} 1 & 2 & 3 & \cdots & n\\ n & 1 & 2 & \cdots & n - 1\\ n - 1 & n & 1 & \cdots & n - 2\\ \cdots & \cdots & \cdots & \cdots & \cdots\\2 & 3 & 4 & \cdots & 1 \end{pmatrix}
\end{tabular}
Show thatϵndet(In−A2n)+ϵn−1det(ϵIn−A2n)+ϵn−2det(ϵ2In−A2n)+⋯+det(ϵnIn−A2n)\epsilon^ndet\left(I_n-A^{2n}\right)+\epsilon^{n-1}det\left(\epsilon I_n-A^{2n}\right)+\epsilon^{n-2}det\left(\epsilon^2 I_n-A^{2n}\right)+\cdots +det\left(\epsilon^n I_n-A^{2n}\right) =n(−1)n−1[nn(n+1)2]2n2−4n(1+(n+1)2n(2n+(−1)n(2nn)))=n(-1)^{n-1}\left[\frac{n^n(n+1)}{2}\right]^{2n^2-4n}\left(1+(n+1)^{2n}\left(2n+(-1)^n{{2n}\choose{n}}\right)\right)where ϵ∈C\R\epsilon \in \mathbb{C}\backslash \mathbb{R}, ϵn+1=1\epsilon^{n+1}=1
linear algebramatrixdeterminantcomplex numbers
Prove this trigonometric inequality

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
Let xx, yy, z>0z > 0 such that x2+y2+z2=3x^2 + y^2 + z^2 = 3. Then x3tan⁡−11x+y3tan⁡−11y+z3tan⁡−11z<π32x^3\tan^{-1}\frac{1}{x}+y^3\tan^{-1}\frac{1}{y}+z^3\tan^{-1}\frac{1}{z}<\frac{\pi \sqrt{3}}{2}
inequalitiestrigonometry
Prove the inequality and find the limit

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
[*] If aa, bb, cc, d>0d > 0, show inequality:(tan⁡−1(ad−bcac+bd))2≥2(1−ac+bd(a2+b2)(c2+d2))\left(\tan^{-1}\left(\frac{ad-bc}{ac+bd}\right)\right)^2\geq 2\left(1-\frac{ac+bd}{\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}}\right) [*] Calculate lim⁡n→∞nα(n−∑k=1nn+k2−k(n2+k2)(n2+(k−1)2))\lim \limits_{n \to \infty}n^{\alpha}\left(n- \sum \limits_{k=1}^n\frac{n^+k^2-k}{\sqrt{\left(n^2+k^2\right)\left(n^2+(k-1)^2\right)}}\right)where α∈R\alpha \in \mathbb{R}
inequalitieslimittrigonometry
Prove this integral inequality of the convex function

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
Let f:(0,+∞)→Rf : (0,+\infty) \to \mathbb{R} a convex function and α,β,γ>0\alpha, \beta, \gamma > 0. Then 16α∫06αf(x)dx + 16β∫06βf(x)dx + 16γ∫06γf(x)dx\frac{1}{6\alpha}\int \limits_0^{6\alpha}f(x)dx\ +\ \frac{1}{6\beta}\int \limits_0^{6\beta}f(x)dx\ +\ \frac{1}{6\gamma}\int \limits_0^{6\gamma}f(x)dx ≥13α+2β+γ∫03α+2β+γf(x)dx + 1α+3β+2γ∫0α+3β+2γf(x)dx \geq \frac{1}{3\alpha +2\beta +\gamma}\int \limits_0^{3\alpha +2\beta +\gamma}f(x)dx\ +\ \frac{1}{\alpha +3\beta +2\gamma}\int \limits_0^{\alpha +3\beta +2\gamma}f(x)dx\ + 12α+β+3γ∫02α+β+3γf(x)dx+\ \frac{1}{2\alpha +\beta +3\gamma}\int \limits_0^{2\alpha +\beta +3\gamma}f(x)dx
integrationinequalitiesconvex functioncalculusfunction
Prove this 3 variable inequality

Source: 2019 Jozsef Wildt International Math Competition

5/19/2020
Let xx, yy, z>0z > 0, λ∈(−∞,0)∪(1,+∞)\lambda \in (-\infty, 0) \cup (1,+\infty) such that x+y+z=1x + y + z = 1. Then∑cycxλyλ∑cyc1(x+y)2λ≥9(14−19∑cyc1(x+1)2)λ\sum \limits_{cyc} x^{\lambda}y^{\lambda}\sum \limits_{cyc}\frac{1}{(x+y)^{2\lambda}}\geq 9\left(\frac{1}{4}-\frac{1}{9}\sum \limits_{cyc}\frac{1}{(x+1)^2} \right)^{\lambda}
inequalities
Prove these two inequalities on heights, inradius and exradius of a tetrahedon

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
In all tetrahedron ABCDABCD holds
[*] (n(n+2))1n∑cyc((ha−r)2(han−rn)(han+2−rn+2))1n≤1r2(n(n+2))^{\frac{1}{n}} \sum \limits_{cyc} \left(\frac{(h_a-r)^2}{(h_a^n-r^n)(h_a^{n+2}-r^{n+2})}\right)^{\frac{1}{n}}\leq \frac{1}{r^2} [*] (n(n+2))1n∑cyc((ra−r)2(ran−rn)(ran+2−rn+2))1n≤1r2(n(n+2))^{\frac{1}{n}} \sum \limits_{cyc} \left(\frac{(r_a-r)^2}{(r_a^n-r^n)(r_a^{n+2}-r^{n+2})}\right)^{\frac{1}{n}}\leq \frac{1}{r^2}
for all n∈N∗n\in \mathbb{N}^*
tetrahedroninequalitiesinradiusexradius3D geometry
Prove that integration of this two functions from a to b is 1

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Let ff, gg, h:[a,b]→Rh : [a, b] \to \mathbb{R}, three integrable functions such that:∫abfgdx=∫abghdx=∫abhfdx=∫abg2dx∫abh2dx=1\int \limits_a^b fgdx=\int \limits_a^bghdx=\int \limits_a^bhfdx=\int \limits_a^bg^2dx\int \limits_a^bh^2dx=1Then∫abg2dx=∫abh2dx=1\int \limits_a^bg^2dx=\int \limits_a^bh^2dx=1
integrationcalculusfunction
Prove that the value of this integration is 0

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Let f:R→Rf : \mathbb{R} \to \mathbb{R} a periodic and continue function with period TT and F:R→RF : \mathbb{R} \to \mathbb{R} antiderivative of ff. Then ∫0T[F(nx)−F(x)−f(x)(n−1)T2]dx=0\int \limits_0^T \left[F(nx)-F(x)-f(x)\frac{(n-1)T}{2}\right]dx=0
integrationcalculusperiodic function