## Secondary and Higher Secondary Marathon

For students of class 11-12 (age 16+)
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### Re: Secondary and Higher Secondary Marathon

Well,Adib notice that $AK=AC$ he said.

Posts: 217
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### Re: Secondary and Higher Secondary Marathon

If we draw a circle centering at $A$ with radius $AC$ as $CD^2=AD.BD$ it holds that $D$ lies on the radical axis of the two circles.And $DK$ is the radical axis.And we now that the radical axis is perpendicular to the line joining the centres of the concerning circles.The result follows.

Tahmid Hasan
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### Re: Secondary and Higher Secondary Marathon

Nadim Ul Abrar wrote:@Tahmid . Love you bro.
Problem $\boxed {33}$
Let $ABC$ be a right triangle ($\angle C=90$),let $CD$ be the hight and let $K$ be a point in the triangle such that $AK=AC$.Prove that the diametar of the curcumcircle of triangle $ABK$ that contains point $A$ is perpendicular to $DK$
Source : Serbia 1990
Construct three circles $\odot ABC,\odot ABK$ and the circle$(\gamma)$ with radius $AC$ with centre $A$.
Let $M$ be the midpoint of $BC$ which means it is the centre of $\odot ABC$.
Let the centre of $\odot ABK$ be $O$. $C \in \gamma \cap \odot ABC, CD \perp AM \Rightarrow D \in$ radical axis of $\gamma, \odot ABC$.
$D \in AB$ which is the radical axis of $\odot ABC,\odot ABK$. So $D$ is the radical centre of $\gamma,\odot ABC,\odot ABK \Rightarrow DK \perp AO$ Q.E.D.
Sobody else post a new problem.
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বড় ভালবাসি তোমায়,মা

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### Re: Secondary and Higher Secondary Marathon

Problem $\boxed {33}$

Let $ABC$ be a right triangle ($\angle C=90$),let $CD$ be the hight and let $K$ be a point in the triangle such that $AK=AC$.Prove that the diametar of the curcumcircle of triangle $ABK$ that contains point $A$ is perpendicular to $DK$

Source : Serbia 1990
Definitions:
$M$, the mid-point of $AB$.
$O$, the center of circle $AKB$
$KD\cap AO=N$

We know $\displaystyle \frac {AK}{AD}=\frac {AC}{AD}=\frac {AB}{AC}=\frac {AB}{AK}$. So $\triangle AKD\sim \triangle ABK$
Hence $\angle AKB=\angle ADK$
So $\angle KDB=180-\angle ADK=180-\angle AKB=\angle AOM$.
Then $D,N,M,O$ concyclic. So $\angle DMO=\angle DNO=90$.
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Posts: 217
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### Re: Secondary and Higher Secondary Marathon

Problem $\boxed{34}$:$ABCD$ is a cyclic quadrilateral.$E$ and $F$ are variable points on sides $AB$ and $CD$ respectively such that $\displaystyle \frac{AE}{BE}=\frac{CF}{DF}$. $\; P$ is a point on the segment $EF$ such that $\displaystyle \frac{EP}{PF}=\frac{AB}{CD}$.Show that $\displaystyle \frac{[APD]}{[BPC]}$ does not depend on the choices of $E$ and $F$.
Source:IMOSL 1998 G2.

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### Re: Secondary and Higher Secondary Marathon

zadid xcalibured wrote:Problem $\boxed{34}$:$ABCD$ is a cyclic quadrilateral.$E$ and $F$ are variable points on sides $AB$ and $CD$ respectively such that $\displaystyle \frac{AE}{BE}=\frac{CF}{DF}$. $\; P$ is a point on the segment $EF$ such that $\displaystyle \frac{EP}{PF}=\frac{AB}{CD}$.Show that $\displaystyle \frac{[APD]}{[BPC]}$ does not depend on the choices of $E$ and $F$.
Source: IMOSL 1998 G2.
Definitions:
Case $AD\parallel BC$ is trivial. So we may assume $AD\cap BC=X$.

Key Idea:
Firstly we'll try to evaluate $\displaystyle \frac{[APD]}{[BPC]}$. Consider a case where $E=A$ and $F=C$. From $\displaystyle \frac{EP}{PF}=\frac{AB}{CD}$, it follows $XP$ is the angle bisector of $\angle AXB$. So $\triangle APD$ and $\triangle BPC$ have same height. Then $\displaystyle \frac{[APD]}{[BPC]}=\frac {AD}{BC}$.

Now it's time to prove.
Look at $\triangle XAB$ and $\triangle XCD$. Now the given condition implies $\angle AXE=\angle CXF$.....(i)
Also we are given $\displaystyle \frac{EP}{PF}=\frac{AB}{CD}$. Now it implies $XP$ bisects $\angle EXF$....(ii)
From (i) & (ii) it easily follows $XP$ bisects $\angle AXB$. So $\triangle APD$ and $\triangle BPC$ have same height. Then $\displaystyle \frac{[APD]}{[BPC]}=\frac {AD}{BC}$.

Guys, these two marathons have become 'বোরিং মরাথন'. Only geometry problems. Let's get tastes of other things.
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Fahim Shahriar
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### Re: Secondary and Higher Secondary Marathon

Problem $\boxed{35}$

Find all $5$-digit natural numbers $n$ such that $n^2$ ends in all the same as $n$.

Name: Fahim Shahriar Shakkhor
Notre Dame College

SANZEED
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### Re: Secondary and Higher Secondary Marathon

Fahim Shahriar wrote:Problem $\boxed{35}$

Find all $5$-digit natural numbers $n$ such that $n^2$ ends in all the same as $n$.

Can you please tell us what you meant by all the same as $n$? If it meas only the last digit,then the last digit of $n$ will be any of $0,1,5,6$. The remaining digits can be chosen independently. Am I correct?
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nafistiham
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### Re: Secondary and Higher Secondary Marathon

SANZEED wrote:
Fahim Shahriar wrote:Problem $\boxed{35}$

Find all $5$-digit natural numbers $n$ such that $n^2$ ends in all the same as $n$.

Can you please tell us what you meant by all the same as $n$? If it meas only the last digit,then the last digit of $n$ will be any of $0,1,5,6$. The remaining digits can be chosen independently. Am I correct?
I think that is correct. And, so, for the $5$ digit number, there are $9,10,10,10,4$ choices for $1^{st}-5^{th}$ digits.
So, there are $9 \times 10^{3} \times 4=36000$ such numbers.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Fahim Shahriar
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### Re: Secondary and Higher Secondary Marathon

No. Notice that I said 'ends in "all the same" as n'.
I meant the last $5$ digits of $n^2$ is $n$.
Suppose, when $n$ is a two digit number, take $76$ as example. $76^2=5776$. Here the last two digits are same as $n$ or $76$.
Now find 5 digit numbers of this property.
Name: Fahim Shahriar Shakkhor
Notre Dame College