Search found 86 matches
- Thu Aug 04, 2016 10:22 am
- Forum: Geometry
- Topic: Russian Olympiad 1996
- Replies: 2
- Views: 3384
Re: Russian Olympiad 1996
Since, $\angle EAF = \angle FDE $ so $ADFE$ is cyclic. So, $\angle FDA + \angle AEF = 180 $ degrees .....(1) If we can prove that $ ADCB$ is a cyclic, then it'll be done. Now, $\angle ABC + \angle ADC $= $(\angle FEA - \angle EAB) $ + $ (\angle FDA + \angle FDC)$ = $(\angle FDA + \angle AEF) +(\angl...
- Thu Aug 04, 2016 9:59 am
- Forum: Geometry
- Topic: Triangle and grids
- Replies: 0
- Views: 2041
Triangle and grids
There is a grid of equilateral triangles with a distance 1 between any two neighboring grid points. An equilateral triangle with side length $n$ lies on the grid so that all of its vertices are grid points, and all of its sides match the grid. Now, let us decompose this equilateral triangle into $n^...
- Fri May 27, 2016 8:04 pm
- Forum: Divisional Math Olympiad
- Topic: Rangamati math olympiad 13
- Replies: 1
- Views: 2367
- Fri May 27, 2016 2:44 am
- Forum: Geometry
- Topic: Chi. TST 2016 Q6 (day 2)
- Replies: 0
- Views: 2080
Chi. TST 2016 Q6 (day 2)
The diagonals of a cyclic quadrilateral $ABCD$ intersect at $P$, and there exist a circle $\Gamma$ tangent to the extensions of $AB,BC,AD,DC$ at $X,Y,Z,T$ respectively. Circle $\Omega$ passes through points $A,B$, and is externally tangent to circle $\Gamma$ at $S$. Prove that $SP\perp ST$.
- Fri May 27, 2016 2:40 am
- Forum: Geometry
- Topic: A Problem of Romanian TST
- Replies: 10
- Views: 7987
- Fri May 27, 2016 2:17 am
- Forum: Geometry
- Topic: China TST 2016 P1
- Replies: 3
- Views: 3433
Re: China TST 2016 P1
Did you solve it?nahin munkar wrote:Use $ homotety$.It may be a useful way to get the solution.
- Fri May 27, 2016 2:13 am
- Forum: Geometry
- Topic: Easy Problem in geo, (cyclic, angle chasing)
- Replies: 6
- Views: 4858
Re: Easy Problem in geo, (cyclic, angle chasing)
Awesome sala!tanmoy wrote:Apply a projective transformation that sends $ABCD$ to a square $A_{1}B_{1}C_{1}D_{1}$.The rest is very easy
- Fri May 27, 2016 2:10 am
- Forum: Geometry
- Topic: Easy Problem in geo, (cyclic, angle chasing)
- Replies: 6
- Views: 4858
Re: Easy Problem in geo, (cyclic, angle chasing)
I have already solved it and given the main clues of my solution.nahin munkar wrote:Here,we can show $AQYR$ & $XQDP$ is cyclic. So,By easy angle chasing, we can show $\angle RQF$ =$\angle FQP$.That's the proof.[here, RG cuts AC at Y & EP cuts BD at X. It's a simple proof.Try to yourself.]. Proved.
- Thu May 19, 2016 1:00 am
- Forum: Geometry
- Topic: Easy Problem in geo, (cyclic, angle chasing)
- Replies: 6
- Views: 4858
Easy Problem in geo, (cyclic, angle chasing)
$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$, the parallel line $EF$ to $AC$ intersect $AB$ at point $E$, line $FG$ parallel to $BD$ intersect $CD$ at $G$. Let the projection of $E$ onto $CD$ be $P$, projection of $F$ onto $DA$ b...
- Thu May 19, 2016 12:22 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 1970
- Replies: 2
- Views: 3188
Re: IMO 1970
Try yourself ;) The primes which can divide this numbers will be 2,3 or 5 because if the prime factor is larger than this three then it can divide only one number. Three number will be odd. At least one number will be the multiple of 3 and also at least one number will be the multiple of 5.Other odd...