IMO shortlist, 1998

G7. ABC is a triangle with \angle ACB = 2 \angle ABC. D is a point on the side BC such that DC = 2 BD. E is a point on the line AD such that D is the midpoint of AE. Show that \angle ECB + 180 = 2 ∠EBC.

 

Solution.  Put D in the center. We can rewrite everything in terms of a - the number corresponding to A and b - the number corresponding to B. Thus C is -2b, E is -a.
The relation between the angles at B and C is easily reduced to the following equation:
\frac{(a-b)^2}{|a-b|^2}\frac{a+2b}{|a+2b|}=\frac{b^3}{|b|^3}.
Squaring and representing modules with conjugates, we obtain
\frac{(a-b)^2}{(\bar a-\bar b)^2}\frac{a+2b}{\bar a+2\bar b}=\frac{b^3}{\bar b^3}.
Clearing the denominator and setting everything to the left we obtain
a^3\bar b^3-3a\bar b^3 b^3-\bar a^3b^3+3\bar a\bar b^2 b^3=0\ (*)
Similarly, the equality we want to proof is equivalent to
\left(\frac{\frac{-b}{|-b|}}{\frac{-a-b}{|-a-b|}}\right)^2=(-1) \left(\frac{\frac{-a+2b}{|-a+2b|}}{\frac{b}{|b|}}\right)
Both of the arguments of the complex numbers on both sides are between 180 and 360, so squaring is equivalent transformation here. As above we obtain
\frac{(a+b)^2}{(\bar a+\bar b)^2}\frac{a-2b}{\bar a-2\bar b}=\frac{b^3}{\bar b^3}.
Everything on the left and expand - we obtain
-a^3\bar b^3+3a\bar b^3 b^3+\bar a^3b^3-3\bar a\bar b^2 b^3=0.
But this is just the same as (*).

Comments

Post a Comment

Popular posts from this blog

Basel problem type sum, proposed by prof. Skordev

Evaluate \sum_{n=1}^{\infty}\frac{1}{2^n n^2}. Proof. The function S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2} is well defined for x\in[-1,1] and thus we need to find S(1/2). Using differentiation, we observe that S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t. If we make change of variables in the latter integral t\to 1-t we obtain S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1) On the other hand, using integration by parts, we obtain S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t. Notice that \displaystyle \lim_{t\to 0}\log t \log(1-t)=0 (\log (1-t)\sim t and then L'Hopital), hence S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2) Plugging x=1/2 and equating (1) to (2) we obtain \int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2. On the ot...
Read more

Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I. The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.  Let (x_n)_{n\ge 2} be a sequence of real numbers, such that x_2>0 and for every n\ge 2 holds  x_{n+1}=-1+\sqrt[n]{1+nx_n}. Prove consecutively that  1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4. \textbf{Proof.}  \textbf{1)} Clearly all the elements of the sequence are positive. The inequality -1+\sqrt[n]{1+nx_n}<x_n is equivalent to (1+nx_n)<(1+x_n)^n, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence 0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2}, and since the right hand side clearly tends to 0 we obtain $\displaystyle\lim_...
Read more

A problem proposed by prof. Babev

Consider the sequence a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t. It could be proven that \lim_{n\to \infty}n a_n=\frac{1}{2}  Now we are going to prove that \lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} and show how to derive all such limits. \textbf{Proof.} Since \displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2} (for any n\in \mathbb{N}) we can rewrite the limit in question as \lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right) =\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right) Changing the variables t\to n+s and using periodicity of sine, the last simplifies to \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s Since for s\in[0,1], $\displaystyle \...
Read more