p1. In the figure, point $P, Q,R,S$ lies on the side of the rectangle $ABCD$. If it is known that the area of the small square is $ 1$ unit, determine the area of the rectangle $ABCD$. p2. Given a quadratic function $f(x) = x^2 + px + q$ where $p$ and $q$ are integers. Suppose $a, b$, and $c$ are distinct integers so that $2^{2020}$ divides $f(a)$, $f(b)$ and $f(c)$, but $2^{1000}$ doesn't divide $b-a$ nor $c-a$. Prove that $2^{1021}$ divides $b-c$. p3. Find all the irrational numbers $x$ such that $x^2 +20x+20$ and $x^3-2020x+1$ both are rational numbers. p4. It is known that triangle $ABC$ is not isosceles with altitudes of $AA_1, BB_1$, and $CC_1$. Suppose $B_A$ and $C_A$ respectively points on $BB_1$ and $CC_1$ so that $A_1B_A$ is perpendicular on $BB_1$ and $A_1C_A$ is perpendicular on $CC_1$. Lines $B_AC_A$ and $BC$ intersect at the point $T_A$ . Define in the same way the points $T_B$ and $T_C$ . Prove that points $T_A, T_B$, and $T_C$ are collinear. p5. In a city, $n$ children take part in a math competition with a total score of non-negative round. Let $k < n$ be a positive integer. Each child $s$: (i) gets $k$ candies for each score point he gets, and (ii) for every other child $t$ whose score is higher than $s$, then $s$ gets $ 1$ candy for each point the difference between the values of $t$ and $s$. After all the candy is distributed, it turns out that no child gets less candy from Badu, and there are $i$ children who get higher scores than Badu. Determine all values of $i$ which may take.
Problem
Source:
Tags: algebra, geometry, combinatorics, number theory, Indonesia Regional MO
04.10.2021 22:50
p1. Points $A(0,0),B(\frac{9}{\sqrt{5}},0),C(\frac{9}{\sqrt{5}},\frac{4}{\sqrt{5}}),D(0,\frac{4}{\sqrt{5}})$. Points $Q(\frac{2}{\sqrt{5}},0),P(0,\frac{1}{\sqrt{5}})$.
05.10.2021 00:47
@above: I got something different (you're literally so close to the correct answer)
17.05.2023 04:15
P2 We know that \[f(a) = a^2 + pa + q\]\[ f(b) = b^2 + pb + q\]\[f(c) = c^2 + pc + q\]Since $f(a)$,$f(b)$,$f(c)$ are all divisible by $2^{2020}$, then $2^{2020}$ also divides $f(a)-f(b)$ and $f(c)-f(a)$ so \[2^{2020} \mid f(a) - f(b) = (a-b)(a+b+p)\]\[2^{2020} \mid f(c) - f(a) = (c-a)(c+a+p)\]Since $2^{1000} \nmid a-b$ and $2^{1000} \nmid c-a$ then $V_2(a-b), V_2(c-a) \leqslant 999$ which implies \[V_2(a+b+p) \geqslant 2020-999 = 1021\]\[V_2(c+a+p) \geqslant 2020-999=1021\]Hence $2^{1021} \mid (a+b+p)$ and $2^{1021} \mid (c+a+p)$ which implies $2^{1021} \mid b-c$. QED