Okay I did way more problems.. it's just I am too lazy to type up. And Ig no more posts before EGMO cause I am too much busy for boards :( bai people
Problem 1.6: Prove that the number of answers for |a_1|+|a_2|+...+|a_k|≤n is equal to the number of answers for |b_1|+|b_2|+...+|b_n|≤k, where a_i,b_i are integers.
Proof: Let f(n,k) be the number of solutions to |a_1|+\dots+|a_n|\le k.
Note that f(n,k) = f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0)] =f(n-1,k)+f(n-1,k-1)+f(n,k-1).
Similarly, we get f(k,n)=f(k-1,n)+f(k-1,n-1)+f(k,n-1).
Now, we can simply induct on a+b and show f(a,b)=f(b,a).
Done!
EGMO 2012 P2
Let n be a positive integer. Find the greatest possible integer m, in terms of n, with the following property: a table with m rows and n columns can be filled with real numbers in such a manner that for any two different rows \left[ {{a_1},{a_2},\ldots,{a_n}}\right] and \left[ {{b_1},{b_2},\ldots,{b_n}} \right] the following holds:\max\left( {\left| {{a_1} - {b_1}} \right|,\left| {{a_2} - {b_2}} \right|,...,\left| {{a_n} - {b_n}} \right|} \right) = 1
Proof: We have the answer as 2^n. We go with induction. For n=1 it is true. Say it is true for n=k. We will show that it is true for n=k+1, the max possible value of m is 2^{k+1}. Suppose not then m>2^{k+1}.
So there exists 3 rows in the m\times k+1 table such that the first k elements are same (by PHP). So we consider the k+1 th column of the three rows ( \{a_i\}, \{b_i\}, \{b_i\}). We have a_i=b_i=c_i for i\in[k]. Since \max(|a_1-b_1|,\dots,|a_{k+1}-b_{k+1}|)=1.
We get |a_{k+1}-b_{k+1}|=1 |b_{k+1}-c_{k+1}|=1 |a_{k+1}-c_{k+1}|=1 which is impossible to happen as a_{k+1},b_{k+1},c_{k+1} are pairwise distinct.
Equality case construction: write 0,1,\ldots 2^n-1 in binary form and with each digit in one cell in each row.
AIME 2019 P6
In a Martian civilization, all logarithms whose bases are not specified are assumed to be base b, for some fixed b \geq 2. A Martian student writes down
3 \log(\sqrt{x}\log x) = 56 \log_{\log (x)}(x) = 54
and finds that this system of equations has a single real number solution x > 1. Find b
Proof: Let \log_bx=c.
So \log_cx=54\implies x=c^{54}. Also 3\log(\sqrt{x}\log x) =56\implies 3\log(\sqrt{x})+3\log(\log x) =56\implies 3c/2+3\log_bc=56.
But 3\frac{log_xc}{\log_xb}=3\times \frac{1}{54}\times c=\frac{c}{18}\implies 3c/2+c/18=56 \implies c=36\implies x=36^{54}\implies b=216.
EGMO 2019 P1
Find all triples (a, b, c) of real numbers such that ab + bc + ca = 1 and
a^2b + c = b^2c + a = c^2a + b.
Proof: We begin with homogenization. We get a^2b+c[ab+bc+ca]=b^2c+a[ab+bc+ca]\implies c^2(b+a)=c(b^2+c^2)
So c=0 or a^2+b^2=c(a+b).
If c=0 then a^2+b=a=b\implies a=b=\pm 1. Note that a,b\ne 0 as ab=1.
If c\ne 0 ( So assume all are non zero). Then c(a+b)=a^2+b^2 a(c+b)=c^2+b^2 b(a+c)=a^2+c^2 \implies 2ab+2bc+2ca=2a^2+2b^2+2c^2 which by am-gm \implies a=b=c but ab+bc+ca=1\implies a=b=c=\pm \frac{1}{\sqrt{3}}.
USAMO 2006 P4
Find all positive integers n such that there are k \geq 2 positive rational numbers a_1, a_2, \ldots, a_k satisfying a_1 + a_2 + \ldots + a_k = a_1 \cdot a_2 \cdots a_k = n.
Proof: Answer is all positive integers except 1,2,3,5.
For n=1,2,3, we use AM-GM, \frac{a_1+\dots+a_k}{k}\ge \sqrt[k]{a_1\dots a_k}\implies n^{k}\ge k^k\cdot n\implies n^{k-1}\ge k^k\implies n\le 3.
For n, 5^{k-1}\ge k^k. So k\le 2. Let a_1+a_2=a_1a_2=5, then solving we get:
a_1,a_2=\frac{5\pm\sqrt5}2which are both irrational.
For n=prime >7, we have \frac{n}{2}, \frac{1}{2}, 1, \dots.
For n= composite, say n=a\cdot b. So consider n=a\cdot b\cdot 1 \cdot 1\dots.
Serbia 2014
Let ABCD be a quadrilateral such that \angle BCA + \angle CAD = 180^{\circ} and \overline{AB} = \overline{AD} + \overline{BC}. Prove that\angle BAC + \angle ACD = \angle CDA
Proof: Define X=AD\cap CB. Since \angle BCA+\angle CAD=180\implies \angle BCA=180-\angle CAD=\angle XAC\implies XA=XC.
Now, consider point B' on AX such that AB'=BC. Note that AB'BC is an isosceles triangle. Also, we have B'A=BC\implies B'D=AB. So \angle BAC=\angle B'CA. So enough to show B'D=B'C. But B'C=AB=B'D. Done!
Bulgaria EGMO TST 2021 Problem 1
On the side AB of a triangle ABC is chosen a point P. Let Q be the midpoint of BC and let CP and AQ intersect at R. If AB + AP = CP, prove that CR = AB.
Proof: We redefine the points. Define A' as the reflection of A over Q and B' be the reflection of B over A. Define R such that CA'=CR and R\in AQ. Define CR\cap AB=P. Note that \angle PRA=\angle CRA'=\angle AA'C=\angle A'AP\implies AP=PR.
But AB'=AB=A'C=RC\implies PB'=PC\implies PA+AB=CP.
And we are done.
Comments
Post a Comment