warm-up problems for national BdMO'14

[Nayeemul Islam Swad]

$$=> 2^{1000}=(2^b)^{2^a}$$.
$$a=[1,3]$$. So,
$$(a,b)= (1,500),(2,250), (3,125)$$.
So value of all $$a+b=881$$

But why don't you count $a = 0?$
I think that's also a valid $a$.

[asif e elahi]

But why don't you count $a = 0?$
I think that's also a valid $a$.

$a$ and $b$ are positive integers.$0$ is not a positive integer.

[asif e elahi]

Problem $11$:Find all positive integers $m$ and $n$ such that $(m^{2}-n^{2})^{2}=16n+1$

[photon]

Solution of 4 :
circumcircle of $\Delta ADB$ cuts $AC$ at $Q$ . $\angle QBD = \angle QAD = 36^o$ , $\angle QBC + \angle CBD = 36^o \Rightarrow \angle QBC = 18^o $ , similarly $\angle QDC = 36^o$ . As $DC $ and $BC $ are angle bisector of $\angle QDP$ and $\angle QBP$ respectively ; $\displaystyle \frac {QB}{PB} = \frac {QC}{CP} =\frac{QB}{PB}$ ; $\displaystyle \frac {QD}{PD}=\frac {QB}{PB}$ . Now ,

$\displaystyle \frac {PB}{PD}=\frac {AB.sin72^o}{AD.sin36^o}=2cos36^o.\frac {AB}{AD}$

$\displaystyle \frac {PB}{PD}=\frac {QB}{QD}=\frac {sin 72^o}{sin36^o}=2cos 36^o$
from last two eqn $\Rightarrow AB=AD$ , $\therefore \angle BDA = \angle ABD = 36^o$ ,
$\angle APD = 180^o - \angle PDA-\angle PAD=180^o-36^o-36^o=72^o$

[photon]

Problem 12.
Let $G$ be the centroid of a triangle $ABC$, and $M$ be the midpoint of $BC$. Let $X$ be on $AB$ and $Y$ on $AC$ such that the points $X$, $Y$, and $G$ are collinear and $XY$ and $BC$ are parallel. Suppose that $XC$ and $GB$ intersect at $Q$ and $YB$ and $GC$ intersect at $P$. Show that triangle $MPQ$ is similar to triangle $ABC$.

APMO-1993

[sowmitra]

Solution to Problem 12:
Lemma:
If $G$ is the centroid of $\triangle ABC$, $E, F$ the mid-points of $AB, AC$, and, $AG \cap EF=S$, then,
$GS:SA=1:3$ and $S$ is the mid-point of $EF$.
Proof of Lemma:
If $M$ is the mid-point of $BC$, then, $G$ is also the centroid of $\triangle MEF$.
$\therefore GS=2GM=4GA \Rightarrow \frac{GS}{GA}=\frac{1}{4} \Rightarrow \frac{GS}{SA}=\frac{1}{3}$
So, $GS:SA=1:3$ $\square$

First,
$\displaystyle \frac{GX}{MB}=\frac{AG}{AM}=\frac{2}{3} \Rightarrow \frac{GX}{2MB}=\frac{1}{3} \Rightarrow \frac{GX}{BC}=\frac{1}{3}$

Now, $\triangle BQC \sim \triangle XQG.$
$\displaystyle \therefore \frac{GQ}{QB}=\frac{GX}{BC}=\frac{1}{3}$, i.e, $GQ:QB=1:3$
So, by the given lemma, $Q$ is the mid-point of $ME$. Similarly $P$ is the mid-point of $MF$.
$\therefore PQ||EF||BC, MP(=MF)||AB$, and, $MQ(=ME)||AC$
So, $\triangle MPQ \sim \triangle ABC$.