olymmath

JMO 1995

Japan Mathematical Olympiad 1995.

Other problems

Problem 1

Let n≥2n\geq 2 be integers and rr be positive integers such that rr is not the multiple of nn, and let gg be the greatest common measure of nn and rr. Prove that ∑i=1n−1{rin} =12(n−g).\sum_{i=1}^{n-1} \left\{\frac{ri}{n} \right\}\ =\frac{1}{2}(n-g). where {x}\left\{x\right\} is the fractional part, that is to say, which means the value by subtracting xx from the maximum integer value which is equal or less than xx.

Problem 2

Find all the non-constant rational function of f(x)f(x) and real numbers aa satisfying {f(x)}2−a=f(x2).\{f(x)\}^2-a=f(x^2). Here a rational function of xx is the equation expressed by the ratio of two polynominals of x.x.

Problem 3

Given a convex pentagon ABCDE.ABCDE. Let S,RS,R be the intersection points of ACAC or ADAD and BEBE respectively, and let the intersection points T,PT,P of CACA or CECE and BDBD respectively.Let QQ be the intersection point of CECE and AD.AD. If all of △ASR, △BTS, △CPT, △DQP,\triangle{ASR},\ \triangle{BTS},\ \triangle{CPT},\ \triangle{DQP}, and △ERQ\triangle{ERQ} have the area of 1,1, then find the area of the following pentagons.

(1) The pentagon PQRST.PQRST. (2) The pentagon ABCDE.ABCDE.

Problem 4

The sequence {a1, a2, ⋯ }\{a_1,\ a_2,\ \cdots\} is defined by a2n=an, a2n+1=(−1)n.a_{2n}=a_n,\ a_{2n+1}=(-1)^n. A point PP moves on the coordinate plane as follows.

(1) Let P0P_0 be the origin, PP moves in a distance of 11 from P0P_0 toward in the positive direction of xx - axis, Denote this point by Pi.P_i.

(2) After PP has moved to Pi,P_i, it turns 90∘90^\circ to the left and moves in a distance of 11 when ai=1,a_i=1, and turns 90∘90^\circ to the right and moves in a distance of 11 when ai=−1.a_i=-1. Denote this point by Pi+1,P_{i+1}, w here i=1,2,⋯ .i=1,2,\cdots. Prove that PP can't pass on the same segment more than two times.

Problem 5

Let k,nk,n be integers such that 1≤k≤n,1\leq k\leq n, a nd let a1, a2, ⋯ , aka_1,\ a_2,\ \cdots,\ a_k be numbers satisfying the following equations.

{a1+⋯⋯+ak=na12+⋯⋯+ak2=n            ⋮a1k+⋯⋯+akk=n\begin{cases} a_1+\cdots\cdots +a_k=n \\ a_1^2+\cdots\cdots +a_k^2=n\\ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \\ a_1^k+\cdots\cdots +a_k^k=n \end{cases}

Prove that (x+a1)(x+a2)⋯(x+ak)=xk+ nC1 xk−1+ nC2 xk−2+⋯+ nCk.(x+a_1)(x+a_2)\cdots(x+a_k)=x^k+\ _n C_1\ x^{k-1}+\ _nC_2\ x^{k-2}+\cdots+\ _nC_k.

where iCj_iC_j is a binomial coefficient which means i⋅(i−1)⋯(i−j+1)j⋅(j−1)⋯2⋅1.\frac{i\cdot (i-1)\cdots(i-j+1)}{j\cdot (j-1)\cdots 2\cdot 1}.