Beginner's Marathon

For students of class 6-8 (age 12 to 14)
Arko Roy
Posts: 2
Joined: Thu Mar 23, 2017 11:24 pm

Re: Beginner's Marathon

Unread post by Arko Roy » Tue Mar 28, 2017 10:26 pm

Mod note: This solution is wrong. See below for correct solution.
Let we are having 3 points, A1, A2, A3. The broken line would be (A1)(A2)(A3). Since there are 2 lines with common vertex A2, so intersection phenomenon is impossible. But if we have 4 points & create a crossed(*) quadrilateral, there would be an intersection between 2 opposite sides of the quadrilateral. So maximum value of n is 3

(*)in জ্যামিতির ২য় পাঠ, there's mentioned that quadrilateral are of 3 types and one of them is crossed, where 2 opposite sides intersects.

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: Beginner's Marathon

Unread post by Absur Khan Siam » Tue Mar 28, 2017 10:53 pm

Problem$\boxed{6}$
Let $ABC$ be an acute triangle. The lines $l_1$ and $l_2$ are perpendicular to $AB$ at the points $A$ and
$B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$
intersect $l_1$ and $l_2$ at the points $E$ and $F$, respectively. If $D$ is the intersection point of the lines
$EF$ and $MC$, prove that
$\angle ADB = \angle EMF$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

Re: Beginner's Marathon

Unread post by Thamim Zahin » Tue Mar 28, 2017 11:05 pm

Mod Note: This solution is incorrect. See below for correct solution.
Arko Roy wrote:Let we are having 3 points, A1, A2, A3. The broken line would be (A1)(A2)(A3). Since there are 2 lines with common vertex A2, so intersection phenomenon is impossible. But if we have 4 points & create a crossed(*) quadrilateral, there would be an intersection between 2 opposite sides of the quadrilateral. So maximum value of n is 3

(*)in জ্যামিতির ২য় পাঠ, there's mentioned that quadrilateral are of 3 types and one of them is crossed, where 2 opposite sides intersects.

Wrong solution. Correct solution :

First we can say that,

if we choose $5$ points(or more) in plane with no three are collinear,
there is a such $4$ points this points are vertices of a convex quadrilateral.

and in quadrangle diognals are intersect

so that $n=4$ is maximal. If we choose 4 points such that this points are vertices of a concave quadrilateral, this is ok.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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ahmedittihad
Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm

Re: Beginner's Marathon

Unread post by ahmedittihad » Wed Mar 29, 2017 2:06 am

Thamim Zahin wrote:
Arko Roy wrote:Let we are having 3 points, A1, A2, A3. The broken line would be (A1)(A2)(A3). Since there are 2 lines with common vertex A2, so intersection phenomenon is impossible. But if we have 4 points & create a crossed(*) quadrilateral, there would be an intersection between 2 opposite sides of the quadrilateral. So maximum value of n is 3

(*)in জ্যামিতির ২য় পাঠ, there's mentioned that quadrilateral are of 3 types and one of them is crossed, where 2 opposite sides intersects.

Wrong solution. Correct solution :

First we can say that,

if we choose $5$ points(or more) in plane with no three are collinear,
there is a such $4$ points this points are vertices of a convex quadrilateral.

and in quadrangle diognals are intersect

so that $n=4$ is maximal. If we choose 4 points such that this points are vertices of a concave quadrilateral, this is ok.
Thamim, you need to prove the lemma that any 5 points must habe atleast one convex quadrilateral.
Hint of the proof, think about convex hulls.
Frankly, my dear, I don't give a damn.

Epshita32
Posts: 37
Joined: Mon Aug 24, 2015 12:34 am

Re: Beginner's Marathon

Unread post by Epshita32 » Wed Mar 29, 2017 2:48 am

ahmedittihad wrote:
Thamim Zahin wrote:
Arko Roy wrote:Let we are having 3 points, A1, A2, A3. The broken line would be (A1)(A2)(A3). Since there are 2 lines with common vertex A2, so intersection phenomenon is impossible. But if we have 4 points & create a crossed(*) quadrilateral, there would be an intersection between 2 opposite sides of the quadrilateral. So maximum value of n is 3

(*)in জ্যামিতির ২য় পাঠ, there's mentioned that quadrilateral are of 3 types and one of them is crossed, where 2 opposite sides intersects.

Wrong solution. Correct solution :

First we can say that,

if we choose $5$ points(or more) in plane with no three are collinear,
there is a such $4$ points this points are vertices of a convex quadrilateral.

and in quadrangle diognals are intersect

so that $n=4$ is maximal. If we choose 4 points such that this points are vertices of a concave quadrilateral, this is ok.
Thamim, you need to prove the lemma that any 5 points must habe atleast one convex quadrilateral.
Hint of the proof, think about convex hulls.
Suppose we have 5 points - A, B, C, D, E. Here, we can take D inside the triangle ABC. But then there is nowhere to put E. If it is in 1, then ABDE is self-intersecting. Similarly, in 2, ACDE is self-intersecting In 3, BCDE is self-intersecting. In 4, ADCE is self intersecting and so on.

So any 5 points must have at least one convex quadrilateral. Then when one takes the arrangement making it a cross quadrilateral, the broken line will intersect.

Edit: Cause if we take any 4 points, their convex hull must be a triangle. i.e. Construction for four points: Take the vertices of a triangle and any point in its interior. None of the six possible line segments intersect, so any possible broken line does not intersect itself.

Note: The convex hull is the smallest convex polygon containing all the points. Think of it as a fence which surrounds all the points.

Mod Note: Added some clarification. All corrections are in this color.
Last edited by Zawadx on Fri Mar 31, 2017 9:03 pm, edited 3 times in total.
Reason: Hiding comments + Explanatory Note

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Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

Re: Beginner's Marathon

Unread post by Thamim Zahin » Wed Mar 29, 2017 11:24 am

I am seeing P7, but nobody solved P6. And one more thing, does everyone hate Number Theory or Algebra? I am not seeing any of this. And i am asking @Zawadx, can i post ineq and FE?
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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Zawadx
Posts: 90
Joined: Fri Dec 28, 2012 8:35 pm

Re: Beginner's Marathon

Unread post by Zawadx » Wed Mar 29, 2017 3:17 pm

Deleted P7 as P6 hasn't yet been solved. Feel free to repost it after a solution has been posted for P6.
Thamim Zahin wrote:I am seeing P7, but nobody solved P6. And one more thing, does everyone hate Number Theory or Algebra? I am not seeing any of this. And i am asking @Zawadx, can i post ineq and FE?
This thread is aimed at beginners, so no you may not post ineq or FE

M Ahsan Al Mahir
Posts: 16
Joined: Wed Aug 10, 2016 1:29 am

Re: Beginner's Marathon

Unread post by M Ahsan Al Mahir » Wed Mar 29, 2017 9:34 pm

Solution to P6:

Let $ME$ meet $AC$ at $X$, $MF$ meet $AB$ at $Y$.
Let the circumcircles of $\triangle CEX$, $\triangle CFY$ meet at $T$. $\angle FYC = \angle FTC = \angle ETC = \angle EXC = 90$. So $E, F, T$ are colliner. As $MX \cdot ME = MA^2 = MB^2 = MY \cdot MF$, $M$ lies on the radical axis of $\odot CFY$ & $\odot CEX$. So $C, T, M$ are collinear.
Let $H$ be the orthocenter of $\triangle ABC$. Let $BH$ meet $EF$ at $H'$. So, $\angle BH'F = \angle MET = \angle MAT$. So $\square ATH'B$ is cyclic. If we let $AH$ meet $EF$ at $H''$ , we can similarly show that $ATH''B$ is cyclic. Which leads us to the conclusion that $H' = H'' = H$. So, $\angle ATB = \angle AHB = \angle FME$. (As $AH \parallel MF$ and $BH \parallel ME$). So we're done.

Epshita32
Posts: 37
Joined: Mon Aug 24, 2015 12:34 am

Re: Beginner's Marathon

Unread post by Epshita32 » Thu Mar 30, 2017 12:16 am

$\text{Problem } 7$

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.

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Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

Re: Beginner's Marathon

Unread post by Thamim Zahin » Thu Mar 30, 2017 12:33 am

$\text{Solution to Problem 7}$

Name all number $a_1,a_2 \cdots a_{2000}$. Where a_i is neighbour to $a_{i-1}$ and $a_{i+1}$.

Lets thing all of them aren't equal. In that case we will have $a_i \not = a_{i+2}$.

WLOG, take $a_i>a_{i+2}$. (There is a other case where all $a_{2j}$ and all $a_{2j+1}$ are equal. But it is clearly impossible. Because the middle will also be equal.)

So, $a_i>a_{i+1}>a_{i+2} [a_i=\frac{a_{i-1}+a_{i+1}}{2}$ and $a_i>a_{i+1} so, a_{i-1}>a_i]$.

From same logic, $a_{i-2}>a_i>a_{i+1}>a_{i+2}$.

If we continue this process, we will have $a_{i-k}>a_{i+1-k}>\cdots > a_i>a_{i+1}>a_{i+2}$. But all of them are on circle.

So, $a_i>a_i$. Contradiction. So, all of them must be equal or, $a_i=a_j$ (where $i,j \le 2000$).
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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