BdMO Online Forum

Epshita32 wrote:P7. There are 2000 points on a circle and each point is given a number which is equal to the average of the two numbers which are its nearest neighbors. Show that all the numbers must be equal.

An alternate Solution:

We use extremal principal. Since there a finite number of numbers, there must exists a minimum number. If the minimal number occurs more than once, randomly pick one. Due to minimality, the numbers on the both sides of those numbers would be equal to the minimal number. This way, all the numbers on the circle will be equal to the minimal number and thus we are done.

Problem 8:

Suppose that 5 points lie on a sphere. Prove that there exists a closed semi-sphere (half a sphere including boundary), which contains 4 of the points.

Solution to P8:
Consider the circle through any two of the points. This partitions the sphere into two hemispheres. By PHP, 2 of the remaining 3 points must lie in one of the hemispheres. These two points and the original two points lie in a closed semi-sphere.

$\text{Problem }9$

There is a billiard ball rolling on a circular table. Everytime it hits the edge, it gets reflected (Assume the ball hits the circle at X . The course of the ball gets reflected with respect to the tangent from X). Prove that, if the ball goes through a point $P$ three times, the ball goes through that point $P$ infinitely many times. Provided that the ball doesn't stop.

Another solution to P6:
As $EM$ is perpendicular on $AC$ we can say by perpendicular lemma,
$CE^2 + AM^2 = AE^2 + CM^2$
or, $CM^2 - AM^2 = CE^2 - AE^2$
Again, $MF$ is perpendicular on $BC$
so, $CF^2 + BM^2 = CM^2 + BF^2$
or, $CM^2 - BM^2 = CF^2 - BF^2 = CM^2 - AM^2$
so, $CE^2 - AE^2 = CF^2 - BF^2$
or, $CE^2 + BF^2 = CF^2 + AE^2$
or, $CE^2 + BF^2 + BM^2 = CF^2 + AE^2 + AM^2$
or, $CE^2 + MF^2 = CF^2 + EM^2$
So, by the reverse of perpendicular lemma, we can say,
$CM$ is perpendicular on $EF$
So, $EAMD$ is cyclic.
$\angle{EMA} = \angle{ADE}$
Again, $MDFB$ is cyclic.
$\angle{BMF} = \angle{BDF}$
So, $\angle{EMF} = \angle{ADB} = 180^o - \angle{EMA} - \angle{BMF} = 180^o - \angle{ADE} - \angle{BDF}$
Proved

So got a bit out of hand. Reducing the max difficulty level to the hardest problems in BDMO. Please don't post problems which ask for theory beyond that presented in the book. If you use advanced ideas, try to explain them so that everyone will understand

Also, let's try to keep this on track with a proper progression of problems and solutions, so that readers can work through them from the start. If you want to post an alternate solution but other problems have already been posted, I recommend starting a new thread. They're very welcome! Also if you must make a comment, please use [*hide] tags so that it doesn't clutter the page. Though if you want to highlight why a solution is wrong (and preferably post the correct solution), do it outside the [*hide] tag.

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ahmedittihad wrote:$\text{Problem }9$

There is a billiard ball rolling on a circular table. Everytime it hits the edge, it gets reflected (Assume the ball hits the circle at X . The course of the ball gets reflected with respect to the tangent from X). Prove that, if the ball goes through a point $P$ three times, the ball goes through that point $P$ infinitely many times. Provided that the ball doesn't stop.

This problem is really cool if you get the gist of it. I'll give some hints. .

ahmedittihad wrote:$\text{Problem }9$

There is a billiard ball rolling on a circular table. Everytime it hits the edge, it gets reflected (Assume the ball hits the circle at X . The course of the ball gets reflected with respect to the tangent from X). Prove that, if the ball goes through a point $P$ three times, the ball goes through that point $P$ infinitely many times. Provided that the ball doesn't stop.

Here's my solution: Most of my solutions in BdMO forum ends with "The rest is trivial".

Problem 10.
Let $S$ be a square of side $2$, and choose $9$ points inside $S$. Show that $3$ of these points may be chosen which are the vertices of a triangle of area $\leq \dfrac{1}{2}$.

Btw, this is a really easy problem and please participate, everyone.

Solution to problem $\boxed{10}$:
We partition the square in $4$ squares with the length of the side $1$.According to PHP, there is a square which contains at least $3$ of the points .We can see that the area of this triangle is smaller than the area of a triangle with the vertices on the sides of the square and then notice that the area of this second triangle is smaller
than the area of a triangle with the vertices in the vertices of the square. But the area of such a triangle is $\dfrac{1}{2}$.
Thus we proof that , area of that triangle is $\leq \dfrac{1}{2}$