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Showing posts from September, 2022

2021 IMO SL #A1

Problem:  Let $n$ be a positive integer. Given is a subset $A$ of $\{0,1,...,5^n\}$ with $4n+2$ elements. Prove that there exist three elements $a<b<c$ from $A$ such that $c+2a>3b$. Solution:  Suppose for the sake of contradiction that $c+2a\le 3b$ for all $a,b,c\in A$ with $a<b<c$.  We claim that if the elements of $A$ are $s_1<s_2<\dots<s_{4n+2}$, then $$s_i\ge s_{4n+2}(1-(\tfrac{2}{3})^{i-1})$$ for all $i$. We use induction to prove this -- the base case is trivially true because $s_1\geq 0$. As for the inductive step, consider $s_k.$ Take $(a,b,c)=(s_{k-1},s_k,s_{4n+2})$; by our assumption, we have $$s_{4n+2}+2s_{k-1}\ge 3s_k.$$ Therefore, $$s_k\ge\frac{s_{4n+2}+2s_{k-1}}{3}\ge\frac{s_{4n+2}+2(s_{4n+2}(1-(\tfrac{2}{3})^{k-2}))}{3}=\frac{s_{4n+2}(3-2(\tfrac{2}{3})^{k-2})}{3}=s_{4n+2}(1-(\tfrac{2}{3})^{k-1}),$$ which completes the induction.  Now, using our claim, we have $$s_{4n+1}\ge s_{4n+2}(1-(\tfrac{2}{3})^{4n})=s_{4n+2}-(\tfrac{16}{81})^{n}s_{4n+2}>s_

2017 USAJMO #3

Let $ABC$ be an equilateral triangle, and point $P$ on its circumcircle. Let $PA$ and $BC$ intersect at $D$, $PB$ and $AC$ intersect at $E$, and $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$. Solution 1:  Note that $$\angle{DPF}=\angle{FPE}=\angle{EPD}=120^{\circ}.$$ Now, using the sin area formula, we get $$[DEF]=[DPF]+[FPE]+[EPD]=\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right).$$We also have that $$[ABC]=BC^2 \cdot \frac{\sqrt{3}}{4}.$$ So, it suffices to prove $$\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right)=2 \cdot \frac{BC^2\sqrt{3}}{4} \implies DP \cdot FP+FP \cdot EP+EP \cdot DP=2BC^2.$$By the Law of Cosines on $\triangle{BPC}$, we obtain $$BC^2=b^2+c^2-2 \cdot b \cdot c \cdot \cos{120^{\circ}}=b^2+c^2+bc,$$ and Ptolemy's Theorem on quadrilateral $ABPC$ gives$$AP \cdot BC=BP \cdot AC+CP \cdot AB \implies AP=b+c.$$From some angle chasing, we know that $\triangle{ACD} \sim \tri

2000 Putnam A1

Problem: Let $A$ be a positive real number. What are the possible values of $$\sum_{j=0}^{\infty} x_j^2,$$ given that $x_0,x_1,\dots$ are positive numbers for which $$\sum_{j=0}^{\infty}=A?$$ Solution:  The answer is $\boxed{(0,A^2)}$. \\ There are two things we need to prove: first, that the possible values of $\sum_{j=0}^{\infty} x_j^2$ must lie in the interval $(0,A^2)$, and second, all values in this interval can be achieved. \\ For our first problem, the values are obviously positive. Also, note that $$\sum_{j=0}^{\infty} \frac{x_j}{A}=1,$$ and because $x_j>0$, we must have $$0 <  \frac{x_j}A   < 1,$$ which implies that $$\left(\frac{x_j}A\right)^2   <  \frac{x_j}A  ,$$$$\implies \sum_{j=0}^{\infty}\left(\frac{x_j}A\right)^2   <  \sum_{j=0}^{\infty}\left(\frac{x_j}A\right)   = 1.$$Multiplying by $A^2$, we get $$\sum_{j=0}^{\infty}x_j^2 < A^2.$$ \\ Now, for the second part, let the sequence of $x_j$'s be a geometric sequence with ratio $r$. Then, we have $$\su

Occasional Putnam problems

Starting now, I will occasionally post Putnam problems on this blog. However, the main content will still be IMO Problems.  These problems will be under the tag #putnam.