Q1. Let ABC be a triangle and let P be an interior point such that <BPC=90°, <BAP=<BCP. Let M,N be the midpoints of AC,BC respectively. Suppose BP=2PM. Prove that A,P,N are collinear
62hhmm.. i checked out "HSBhatt" on goiit... (I dont visit that one very often... infact rarely..)
Was amazed to see ur solutions.... You are Good man :)
62@Prophet: okie.. that is cool... and to remember so much after so long is commendable.. :)
@ashish.. check that one.. you have not used the equality .. but you have put the LHS Of it as 0... That is what i meant...
106cudnt understand wat u r trying to say.
This is what i did.
I first expanded [x2 + 2x] and got an expression
[x2 + 2x] = [x2] + 2[x] + [2[x]{x} + {x}2 + 2{x}]
Now it is given that
[x2 + 2x] = [x2] + 2[x]
So, [2[x]{x} + {x}2 + 2{x}] = 0
i.e. 0<= 2[x]{x} + {x}2 + 2{x}] <1
62sorry dude.. my wrong.. let me read and try to find the mistake again... :)
24hey .......I want to clear something here. I am not at goiit.com....Maybe "theprophet or I should say hsbhatt" is mistaken..
1I saw goiit , not bad but many irrelevant posts .
Nice having you here , prophet !! Great reply too !
eureka ,
Had prophet not said about himself (some hsbhatt) , would you have known about him ? Internet is a real tricky place .
9great
sir
prophet u must make urself known to us that u r a senior teacher . I till now thought u were an exceptionally gifted student who finished a lot of mathematical work within class 12 ;)
i thank ur service for us both as prophet and hs batt ;)
9and diptosh sir hsbhatt is a prominent expert in goiit as far as i know
cos hes well known to jee aspirants like me ;)
341As i said i wanted to have some fun for a while. btw, celestine did you write INMO?
1haha so theres a secret broken [4]
i smelled something fishy in the name prophet itself
hehe by the way thanx prophet (bhaiyya [3])
95) take A as smallest angle
u can easily prv that
hmax = 4RsinBsinC
AH= 2RcosA and so on
eq reduces to Σ cosA ≤ 2sinBsinC
cos B + cos C≤ cos B-C
cosB + cosC ≤ 1
sinA/2cosB-C/2≤1/2
now A≤60
so sinA/2≤ 1/2
so abv inequality also true
6) same way as prophet
but i din take ab = q
took x+y= p
and x2+y2 -xy = 1/p
rewrote the inequality in forms of p to get
(p-1)2(p-8/5)≤0 ............1
now using ap≥ gp
xy≤(1/2)2/3
also p =1 + 3xy/( p +1 +1/p)
p≤1 + 3(1/2)2/3/( p +1 +1/p)
p≤1 + (1/2)2/3 as p + 1/p ≥2
now i prove that
1 + (1/2)2/3 ≤ 8/5 so that the abv eq1 is satisfied always
(1/2)2/3 ≤3/5
125≤128 which is true hence proved
9no sir wasnt selected (i donno why even though i got 5 out of 6 correct includin 2 one in rmo which was toughest ;(
ne way solved inmo paper for fun
1 ) simple but long . use sine rule and properties of triangle to prove that
<ANB = <PNB collinearity follows
2 ) its of frm Σ 10-r2 were r goes frm 1 to infinty . its obvious it cant be rational cos its continuing aperiodically but i donno how to give abs mathematical proof
3)asish got ans as [n,√1+(1+n)2 -1)) ur method is fine
4) u can prove that a isoscles or equlateral triangle must exist . the other condition as i see it is jus to mislead u in my view ,but u cant say nething in inmo
proof is by taking R,B,G as 3 colours
WLOG take 2 pts of R and construct a circle with one pt as centre. now no pt on circle can be R ( else a isoΔ frms )
take 2 pts B on circle and construct 2 more circles taking each as centre once . now the 2pts of intersection of these circles on main circle has to be G. now cosider pt midway of those two Gs on circle its also pt midway of those Bs . that pt cant be R,B,G else a isoΔ follows.
hence proof that isoΔ exists ;)
1i gave INMO too n think almost all of my solutions wer dif....
@celestine....frn the thing is that i talked to our reginal coordinator and he said that many a times students are not given marks 4 methods other than those decided by the paper settin committe....me thinks this isnt fair
106yup celestine even i got that answer for Q3.
By the way u r selected for INCHO right??
9oh sid thats stupidity most of my methods invlve symmetry and analogy
moreover the level of inmo paper is too easy . i solved it in 2 hr easily. compared to fiitjees aits this is nothing . they shud make inmo an exam of elimination rather than selection ;(all my views are biased cos im frustrated by the fact i cudnt clear rmo itself ;(
9btw in q 4 why did they give that angles in gp fact ?
wasnt neccessary right or am i rong somewhere
9see the INMO official solution for q4
Its quite narrow minded i think
15] my way 1)
let O be the circumcenter of ABC, and D, E, F the midpoints of BC, CA, AB;
then AH = 2 * OD and symmetrically,
so it remains to prove that OD + OE + OF <= max altitude of ABC,
but this follows from the fact that the sum of the distances from a point P to the sides of triangle ABC is linear with respect to P, and thus takes its maximum at a vertex of ABC
15] my way 2 one line soln
HBC/ABC=HD/AD
ABHC/ABC=AH/HD that is the ratio of areas wat i ahev written. then symmetrically add up. u will get. thats done.
here D is foot of perpendicular from A to BC
106Q3. Find all real numbers x such that
[x2 +2x] = [x]2 + 2[x]
(Here [x] denotes the largest integer not exceeding x)
106Q4. All the points in the plane are coloured using three colours. Prove that there exists a triangle with vertices having the same colour such that either it is isosceles or its angles are in geometric progression.
106Q5. Let ABC be an acute-angled triangle and let H be its orthocentre. Let hmax denote the largest altitude of the triangle ABC. Prove that
AH + BH + CH <= 2hmax