Problem
Prove that $1/1999<\ln (1999/1998 )<1/1998 $
Solution
Since $1999/1998=1+1/1998$, by the power series expansion of $\ln(1+x)$ we have that $\ln(1999/1998)<1/1998$. Also, since $\ln(1999/1998)=-\ln(1998/1999)=-\ln(1-1/1999)$ and $\ln(1-1/1999)<-1/1999$, we have that $1/1999<\ln(1999/1998)$.
Friday, December 2, 2011
Thursday, December 1, 2011
Thursday, December 1st
Problem
Find
$S=\sum_{i=0}^{101} \frac{x_i^3}{1-3x_i+3x_i^2}$
where $x_i=i/101$.
Solution
Since
$\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{(1-x_i)^3+x_i^3}$
is symmetric when exchanging $x_i$ by $1-x_i$, we have that every such pair will contribute to the sum by one, and there are 51 such pairs, hence $S=51$.
Find
$S=\sum_{i=0}^{101} \frac{x_i^3}{1-3x_i+3x_i^2}$
where $x_i=i/101$.
Solution
Since
$\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{(1-x_i)^3+x_i^3}$
is symmetric when exchanging $x_i$ by $1-x_i$, we have that every such pair will contribute to the sum by one, and there are 51 such pairs, hence $S=51$.
Wednesday, November 30, 2011
Wednesday, November 30th
Problem
Find all naturals $a, b$ such that $\frac{a+1}{b}$ and $\frac{b+1}{a}$ are naturals.
Solution
The only possibilities are $a=b$ and $b=a+1$ (or $a=b+1$). Thus the only solutions are $(1,1)$, $(1,2)$ and $(2,3)$.
Find all naturals $a, b$ such that $\frac{a+1}{b}$ and $\frac{b+1}{a}$ are naturals.
Solution
The only possibilities are $a=b$ and $b=a+1$ (or $a=b+1$). Thus the only solutions are $(1,1)$, $(1,2)$ and $(2,3)$.
Tuesday, November 29th
Problem
Find all the possible areas on an hexagon with equal angles and whose sides are 1,2,3,4,5, and 6 in some order.
Solution
Up to cyclic permutations and reflexions, the only possibilities for the sides are 1,6,2,4,3,5 and 1,6,3,2,5,4. Thus the possible values of areas are $\frac{67\sqrt{3}}{4}$ and $\frac{65\sqrt{3}}{4}$.
Find all the possible areas on an hexagon with equal angles and whose sides are 1,2,3,4,5, and 6 in some order.
Solution
Up to cyclic permutations and reflexions, the only possibilities for the sides are 1,6,2,4,3,5 and 1,6,3,2,5,4. Thus the possible values of areas are $\frac{67\sqrt{3}}{4}$ and $\frac{65\sqrt{3}}{4}$.
Monday, November 28th
Problem
Let $E$ be an ellipse and $E'$ be its reflection along one of its tangents. Find the locus of the foci of $E'$ as the tangent line to $E$ varies.
Solution
Let $A$ and $B$ be the foci of $E$. Then by considering the reflexions of $A$ and $B$ over the tangent lines to $E$ one can see that the locus of $A$ is homothetic to the original ellipse with center at $B$ and a factor of 2. Likewise the locus of $B$ is homothetic to $E$ with center at $A$ and a factor of 2.
Let $E$ be an ellipse and $E'$ be its reflection along one of its tangents. Find the locus of the foci of $E'$ as the tangent line to $E$ varies.
Solution
Let $A$ and $B$ be the foci of $E$. Then by considering the reflexions of $A$ and $B$ over the tangent lines to $E$ one can see that the locus of $A$ is homothetic to the original ellipse with center at $B$ and a factor of 2. Likewise the locus of $B$ is homothetic to $E$ with center at $A$ and a factor of 2.
Tuesday, November 22nd
Problem
Let ABCD be a square and F be a point on BC; draw the perpendicular to DF through B and let it cut DC at Q. Find the angle FQC.
Solution
Let P be intersection BQ with DF. Since the quadrilaterals CQPF and CDBQ are both cyclic, we have that the angles FQC, FPC and DBC are equal we have that FQC is 45.
Let ABCD be a square and F be a point on BC; draw the perpendicular to DF through B and let it cut DC at Q. Find the angle FQC.
Solution
Let P be intersection BQ with DF. Since the quadrilaterals CQPF and CDBQ are both cyclic, we have that the angles FQC, FPC and DBC are equal we have that FQC is 45.
Tuesday, November 22, 2011
Monday, November 21st
Problem
Let $f(x), g(x)$ be two continuous real functions such that $\int_0^1f^2(x)dx=\int_0^1 g^2(x)dx=1$. Prove that there is a real number $c$ such that $f(c)+g(c)\le 2$.
Solution
Suppose that $f(c)+g(c) > 2$ for all $c\in[0,1]$, then $\int_0^1 f(x)g(x)dx>1$, but by Cauchy-Schwarz $\int_0^1 f(x)g(x)dx\le1$ which gives a contradiction, hence such $c$ exists.
Let $f(x), g(x)$ be two continuous real functions such that $\int_0^1f^2(x)dx=\int_0^1 g^2(x)dx=1$. Prove that there is a real number $c$ such that $f(c)+g(c)\le 2$.
Solution
Suppose that $f(c)+g(c) > 2$ for all $c\in[0,1]$, then $\int_0^1 f(x)g(x)dx>1$, but by Cauchy-Schwarz $\int_0^1 f(x)g(x)dx\le1$ which gives a contradiction, hence such $c$ exists.
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