Search found 81 matches

by nahin munkar
Mon Jan 09, 2017 2:59 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63396

Re: Geometry Marathon : Season 3

Now, an easy.problem :D Problem 12: Let $\triangle ABC$ be scalene, with $BC$ as the largest side. Let $D$ be the foot of the perpendicular from $A$ on side $BC$. Let points $K,L$ be chosen on the lines $AB$ and $AC$ respectively, such that $D$ is the midpoint of segment $KL$. Prove that the points ...
by nahin munkar
Mon Jan 09, 2017 12:15 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63396

Re: Geometry Marathon : Season 3

Problem 11 : Let $ABC$ be a triangle inscribed circle $(O)$, orthocenter $H$. $E,F$ lie on $(O)$ such that $EF\parallel BC$. $D$ is midpoint of $HE$. The line passing though $O$ and parallel to $AF$ cuts $AB$ at $G$. Prove that $DG\perp DC$. Solution of problem 11 : We first denote some extra point...
by nahin munkar
Fri Jan 06, 2017 8:59 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63396

Re: Geometry Marathon : Season 3

Problem 4: Let $ABC$ be a triangle and $m$ a line which intersects the sides $AB$ and $AC$ at interior points $D$ and $F$, respectively, and intersects the line $BC$ at a point $E$ such that $C$ lies between $B$ and $E$. The parallel lines from the points $A$, $B$, $C$ to the line $m$ intersect the...
by nahin munkar
Fri Jan 06, 2017 8:19 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63396

Re: Geometry Marathon : Season 3

$\text{Problem 3:}$ In Acute angled triangle $ABC$, let $D$ be the point where $A$ angle bisector meets $BC$. The perpendicular from $B$ to $AD$ meets the circumcircle of $ABD$ at $E$. If $O$ is the circumcentre of triangle $ABC$ then prove that $A,E$ and $O$ are collinear. Solution of problem 3 : ...
by nahin munkar
Fri Jan 06, 2017 7:16 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63396

Re: Geometry Marathon : Season 3

Problem 2 In $\triangle ABC$, $\angle ABC=90^{\circ}$. Let $D$ be any point on side $AC$, $D \neq A,C$. The circumcircle of $\triangle BDC$ and the circle with center $C$ and radius $CD$ intersect at $D,E$. Let $F$ be a point on side $BC$ so that $AF \parallel DE$. $X$ is another point on $BC$(Diff...
by nahin munkar
Thu Jan 05, 2017 11:42 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63396

Geometry Marathon : Season 3

$\Re$evived $\Re$ules : Let's revive geo marathon (after 6 yrs only :mrgreen: ) . The rules will be almost same as before, just enhance solving time duration for 2 days . The difficulty level should be around G1-G5 compared with ISL(IMO Shortlist). Solver will post his own solution of the former pr...
by nahin munkar
Thu Jan 05, 2017 10:16 pm
Forum: Geometry
Topic: GEOMETRY MARATHON: SEASON 2
Replies: 11
Views: 5443

Re: GEOMETRY MARATHON

Next thread goes to the following page :

viewtopic.php?f=25&t=3802
by nahin munkar
Thu Jan 05, 2017 7:29 pm
Forum: Geometry
Topic: GEOMETRY MARATHON: SEASON 2
Replies: 11
Views: 5443

Re: GEOMETRY MARATHON: SEASON 2

New Problem! Let $BE$ and $CF$ be the altitudes in an acute triangle $ABC$. Two circles passing through the points $A$ and $F$ are tangent to the line $BC$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that the lines $PE$ and $QF$ intersect on the circumcircle of $\triangle ...
by nahin munkar
Mon Dec 12, 2016 1:45 pm
Forum: Secondary Level
Topic: Geometry PROBLEM/
Replies: 1
Views: 1684

Re: Geometry PROBLEM/

Use the formula : $ \frac{(n-2)*180^0}{n} $ for measuring each angle of a regular $ n $-gon. Now, have a try. :)
Solution :
$ \angle A$ = $360^0$ - $(108^0 + 120^0)$= $132^0$ :idea:
by nahin munkar
Thu Dec 08, 2016 7:48 pm
Forum: Secondary Level
Topic: Points contained in a bounded area
Replies: 1
Views: 2159

Re: Points contained in a bounded area

Let , within $n$ points, $\triangle XYZ$ is a triangle formed by such $3$ points where the area is maximal. Here,surely, $[XYZ] \le 1$. Now, we construct a triangle $\triangle PQR$ whose medial triangle is $\triangle XYZ$. So, now, $[PQR]=4[XYZ] \le 4$. We claim , all the points are contained in $\...