Subcontests
(6)construct regular heptadecagon PLEASETELLMEYOUARENOTDRAWINGTHIS
Let b≥2 and w≥2 be fixed integers, and n=b+w. Given are 2b identical black rods and 2w identical white rods, each of side length 1.We assemble a regular 2n−gon using these rods so that parallel sides are the same color. Then, a convex 2b-gon B is formed by translating the black rods, and a convex 2w-gon W is formed by translating the white rods. An example of one way of doing the assembly when b=3 and w=2 is shown below, as well as the resulting polygons B and W.[asy]size(10cm);
real w = 2*Sin(18);
real h = 0.10 * w;
real d = 0.33 * h;
picture wht;
picture blk;draw(wht, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle);
fill(blk, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle, black);// draw(unitcircle, blue+dotted);// Original polygon
add(shift(dir(108))*blk);
add(shift(dir(72))*rotate(324)*blk);
add(shift(dir(36))*rotate(288)*wht);
add(shift(dir(0))*rotate(252)*blk);
add(shift(dir(324))*rotate(216)*wht);add(shift(dir(288))*rotate(180)*blk);
add(shift(dir(252))*rotate(144)*blk);
add(shift(dir(216))*rotate(108)*wht);
add(shift(dir(180))*rotate(72)*blk);
add(shift(dir(144))*rotate(36)*wht);// White shifted
real Wk = 1.2;
pair W1 = (1.8,0.1);
pair W2 = W1 + w*dir(36);
pair W3 = W2 + w*dir(108);
pair W4 = W3 + w*dir(216);
path Wgon = W1--W2--W3--W4--cycle;
draw(Wgon);
pair WO = (W1+W3)/2;
transform Wt = shift(WO)*scale(Wk)*shift(-WO);
draw(Wt * Wgon);
label("W", WO);
/*
draw(W1--Wt*W1);
draw(W2--Wt*W2);
draw(W3--Wt*W3);
draw(W4--Wt*W4);
*/// Black shifted
real Bk = 1.10;
pair B1 = (1.5,-0.1);
pair B2 = B1 + w*dir(0);
pair B3 = B2 + w*dir(324);
pair B4 = B3 + w*dir(252);
pair B5 = B4 + w*dir(180);
pair B6 = B5 + w*dir(144);
path Bgon = B1--B2--B3--B4--B5--B6--cycle;
pair BO = (B1+B4)/2;
transform Bt = shift(BO)*scale(Bk)*shift(-BO);
fill(Bt * Bgon, black);
fill(Bgon, white);
label("B", BO);[/asy]Prove that the difference of the areas of B and W depends only on the numbers b and w, and not on how the 2n-gon was assembled.Proposed by Ankan Bhattacharya But How...
Let a0,b0,c0 be complex numbers, and define \begin{align*}a_{n+1} &= a_n^2 + 2b_nc_n \\ b_{n+1} &= b_n^2 + 2c_na_n \\ c_{n+1} &= c_n^2 + 2a_nb_n\end{align*}for all nonnegative integers n. Suppose that max{∣an∣,∣bn∣,∣cn∣}≤2022 for all n. Prove that ∣a0∣2+∣b0∣2+∣c0∣2≤1. Infinite Sequences of Integers
For which positive integers m does there exist an infinite arithmetic sequence of integers a1,a2,... and an infinite geometric sequence of integers g1,g2,... satisfying the following properties?[*] an−gn is divisible by m for all integers n≥1;
[*] a2−a1 is not divisible by m.Holden Mui