Search found 244 matches

by Nadim Ul Abrar
Tue Jan 22, 2013 8:14 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 63710

Re: IMO Marathon

$\boxed {20}$ Let perpendicular bisector of $BC$ intersect circle $(ABC)$ at $T$ and $F$ , Its wellknown that $A,I,F$ are colinear . Now $AE||BC$ imply $TF \perp AE$ , Again $IE \perp AE$ So $IE || TF$ Now $\angle PAI=\angle PAF=\angle PTF=\angle PEI$ imply $A,P,I,E$ cyclic . So $API=90$ . Note that...
by Nadim Ul Abrar
Tue Jan 22, 2013 12:00 am
Forum: Combinatorics
Topic: Comb
Replies: 1
Views: 1671

Comb

$22$ points are chosen from the $7×7$ grid of points $(i,j)$, where $1≤ i≤ 7$
and $1≤ j≤ 7$. Prove that four of the chosen points are the vertices
of a rectahgle with horizontal and vertical sides. Show that this might
not be the case, if $21$ vertices are chosen.
by Nadim Ul Abrar
Mon Jan 21, 2013 11:51 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 63710

Re: IMO Marathon

Problem $\boxed {19 }$ Points $C,M,D$ and $A$ lie on line $l $ in that order with $CM = MD$. Circle $\omega $ is tangent to line at $A$. Let $B$ be the point on $\omega$ that is diametrically opposite to $A$. Lines $BC$ and $BD$ meet $\omega$ at $P$ and $Q$. Prove that the lines tangent to $\omega$ ...
by Nadim Ul Abrar
Mon Jan 21, 2013 11:40 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58436

Re: Secondary and Higher Secondary Marathon

@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 1...
by Nadim Ul Abrar
Sat Jan 19, 2013 9:01 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58436

Re: Secondary and Higher Secondary Marathon

Fun: For problem 28 , EFGH is রম্বস as well .

Problem 29 : Find all positive integers $n$ and $k$ such that $(n+1)^n=2n^k+3n+1$
(spain '12 D2 1)
by Nadim Ul Abrar
Sat Jan 19, 2013 8:32 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58436

Re: Secondary and Higher Secondary Marathon

28 . $\displaystyle YA.YD=YB.YC \rightarrow \frac {YA}{YB}=\frac{YC}{YD} \rightarrow \frac {AG}{BG}=\frac{CH}{DH}$ Similarly , $\displaystyle \frac {AE}{DE}=\frac{CF}{BF}$ Using sine rule $\displaystyle \frac {YA}{YB}=\frac{XA}{XB}=\frac {sin \angle YAB}{sin \angle YBA} \rightarrow \frac {AG}{BG}=\f...
by Nadim Ul Abrar
Fri Jan 11, 2013 11:19 pm
Forum: Geometry
Topic: SD=SM
Replies: 5
Views: 2299

Re: SD=SM

Here we go
SD+SM.PNG
SD+SM.PNG (30.7 KiB) Viewed 2285 times
by Nadim Ul Abrar
Fri Jan 11, 2013 10:45 pm
Forum: Geometry
Topic: IMO 2007 Problem 4
Replies: 4
Views: 8909

Re: IMO 2007 Problem 4

My proof : $\displaystyle \frac{1}{2}. RC .\frac{1}{2}BC.sin(\frac{c}{2})-\frac{1}{2} .(\frac{1}{2}BC)^2.tan(\frac{c}{2})=\frac{1}{2}. RC. \frac{1}{2}AC.sin(\frac{c}{2})-\frac{1}{2} .(\frac{1}{2}AC)^2.tan(\frac{c}{2})$ $ \longleftrightarrow \displaystyle RC.cos(\frac{c}{2})=\frac{AB+AC}{2}$ if $AB \...
by Nadim Ul Abrar
Fri Jan 11, 2013 10:25 pm
Forum: Geometry
Topic: SD=SM
Replies: 5
Views: 2299

SD=SM

In triangle $ABC$ , Let $D$ be the foot of perpendicular from $A$ on $BC$ and $M$ be the midpoint of $BC$ . Points $P,Q$ are on line $AB,AC$ so that $AP=AQ$ and $PQ$ goes through $M$ . Let $S$ be the circumcenter of triangle $APQ$ .Prove that $SD=SM$
by Nadim Ul Abrar
Tue Jan 01, 2013 7:05 pm
Forum: Algebra
Topic: APMO ! How Nice
Replies: 5
Views: 2311

APMO ! How Nice

Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$\sqrt{a+b−c}+\sqrt{b+c−a}+\sqrt{c+a−b} \leq \sqrt{a}+\sqrt{b}+\sqrt{c}$.

(1996)