Search found 186 matches

by photon
Fri Oct 26, 2012 11:34 am
Forum: Geometry
Topic: Cyclic Quad
Replies: 2
Views: 1072

Re: Cyclic Quad

let $OM,ON$ intersect $CD,AB$ at $N_1,M_1$. in $\Delta OCD$ , $ON$ is a median.$\displaystyle \frac{OC}{OD}=\frac{sin\angle NOD}{sin\angle NOC}=\frac{sin\angle BOM_1}{sin\angle AOM_1}=\frac{BM_1}{OB}\frac{OA}{AM_1}$ $\therefore \displaystyle \frac{OC}{OD}=\frac{OA.BM_1}{OB.AM_1}....... (1)$ ,similar...
by photon
Tue Oct 23, 2012 9:21 pm
Forum: Secondary Level
Topic: Seems easy
Replies: 4
Views: 1479

Re: Seems easy

let $O$ is the intersecting point , ${\frac {OA}{OA_1}}={\frac {\Delta AOB}{\Delta A_1OB}}={\frac {2\Delta AOC_1}{2\Delta A_1OC}}={\frac {\frac {1}{2}. OA.OC_1 sin \angle AOC_1}{\frac {1}{2}.OA_1.OCsin \angle A_1OC}}={\frac {OA.OC_1}{OA_1.OC}}$ $\Rightarrow OC_1=OC$ ; $\therefore \Delta BOC_1=\Delta...
by photon
Mon Oct 22, 2012 8:11 pm
Forum: Higher Secondary Level
Topic: perimeter of an isosceles tri.
Replies: 1
Views: 3640

perimeter of an isosceles tri.

in an isosceles triangle $ABC$ ( $AB=AC$), $M$ in a point on $BC$ such that $BM=CM$ .$\angle BXC=3\angle BAC$ where $X$ is a point on $AM$ segment. if , $AX=10,AM=11$ , find the perimeter of the triangle .
by photon
Fri Sep 14, 2012 8:12 pm
Forum: Geometry
Topic: SAMO 2012 Problem 2
Replies: 4
Views: 1473

Re: SAMO 2012 Problem 2

let,YZ=a.height of trapizoid $AYZB$ is $1$. Area($AYZB$)=$\frac{1}{2}.1.(YZ+AB)=\frac{a+1}{2}$ Area($AXB$)=$XYZ+AYZB=\frac{2}{3}+\frac{a+1}{2}=\frac{3a+7}{6}$ . in similar triangle $XYZ,AXB$ $\frac{a^2}{1}=\frac{XYZ}{AXB}=\frac{\frac{2}{3}}{\frac{3a+7}{6}}$ , $\Rightarrow 3a^3+7a^2-4=0 \Rightarrow (...
by photon
Sat Aug 18, 2012 12:24 pm
Forum: Geometry
Topic: AIMO-5 2010 problem 1
Replies: 2
Views: 1008

Re: AIMO-5 2010 problem 1

my solution seems similar to Sanzeed's , but slight different for pic i guess.
by photon
Tue Jun 12, 2012 4:42 pm
Forum: Number Theory
Topic: $x^5-x=k^2$
Replies: 4
Views: 1290


find all positive integer $x$ with proof such that $x^5-x$ is a perfect square .
by photon
Sun May 27, 2012 11:29 am
Forum: Social Lounge
Topic: অভিনন্দন !
Replies: 12
Views: 4566

Re: অভিনন্দন !

congratulation for all!!!! no wish for luck just 10 gallons felix felicis for each one ..... :D :)
by photon
Wed May 23, 2012 3:42 pm
Forum: National Math Camp
Topic: geometry might be easy
Replies: 2
Views: 1360

Re: geometry might be easy

let $BP$ intersects $CA$ at $X$.by angle bisecting theorem, $\frac{CB}{AB}=\frac{CX}{AX}\Rightarrow \frac{CB}{DB}=\frac{CX}{AX}$ so ,$BP,AD$ are parallel. $\angle BPD=\angle ADP,\angle ADC=\angle PBC=\angle ABP$ In $\Delta ADC$ , $\Delta ADM,\Delta MDC$ have same area . $ \frac{1}{2}CD.DM.sin\angle ...
by photon
Tue May 22, 2012 9:02 pm
Forum: Number Theory
Topic: $x^3+x+1$=square
Replies: 4
Views: 1504

Re: $x^3+x+1$=square

afif mansib ch wrote: let\[x=n+1,x^2+1=n-1\]........
it is not true , $a,b$ are co-prime and $ab=pq$ that doesn't mean a=p or q
by photon
Tue May 22, 2012 12:39 pm
Forum: Geometry
Topic: Let's do it together
Replies: 4
Views: 1134

Re: Let's do it together

by ptolemy in cyclic quad $BEPD$, $BE.PD+BD.PE=BP.DE$ $\Rightarrow AC(PD+PE)=BP.DE$ $\Rightarrow \frac{BP}{BD}=\frac{PD +PE}{DE}$ now, $CQ=AC\frac{sin\angle CAQ}{sin\angle AQC}$ $AQ=AC\frac{sin\angle ACQ}{sin\angle AQC}$ \[AQ+CQ=AC(\frac{sin\angle ACQ}{sin\angle AQC}+\frac{sin\angle CAQ}{sin\angle A...