Difference between revisions of "2000 IMO Problems/Problem 1"
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Now, <math>\angle{ABM} = \angle{BMD}</math> since <math>CD</math> is parallel to <math>AB</math>. But <math>AB</math> is tangent to the circumcircle of <math>\triangle{BMD}</math> hence <math>\angle{ABM} = \angle{BDM}</math> and that implies <math>\angle{BMD} = \angle{BDM} . </math>So<math>\triangle{BMD}</math> is isosceles and <math>BM=BD</math>. | Now, <math>\angle{ABM} = \angle{BMD}</math> since <math>CD</math> is parallel to <math>AB</math>. But <math>AB</math> is tangent to the circumcircle of <math>\triangle{BMD}</math> hence <math>\angle{ABM} = \angle{BDM}</math> and that implies <math>\angle{BMD} = \angle{BDM} . </math>So<math>\triangle{BMD}</math> is isosceles and <math>BM=BD</math>. | ||
− | By simple parallel line rules, <math>\angle{ | + | By simple parallel line rules, <math>\angle{EBA}=\angle{BDM}</math>=\angle{ABM}<math>. Similarly, </math>\angle{BAM}=\angle{EAB}<math>, so by </math>\textit{ASA}<math> criterion, </math>\triangle{ABM}<math> and </math>\triangle{ABE}<math> are congruent. |
− | We know that <math>BE=BM=BD< | + | We know that </math>BE=BM=BD<math> so a circle with diameter </math>ED<math> can be circumscribed around </math>\triangle{EMD}<math>. Join points </math>E<math> and </math>M<math>, </math>EM<math> is perpendicular on </math>PQ<math>, previously we proved </math>MP = MQ<math>, hence </math>\triangle{EPQ}<math> is isoscles and </math>EP = EQ$ . |
==See Also== | ==See Also== | ||
{{IMO box|year=2000|before=First Question|num-a=2}} | {{IMO box|year=2000|before=First Question|num-a=2}} |
Revision as of 14:12, 21 June 2024
Problem
Two circles and
intersect at two points
and
. Let
be the line tangent to these circles at
and
, respectively, so that
lies closer to
than
. Let
be the line parallel to
and passing through the point
, with
on
and
on
. Lines
and
meet at
; lines
and
meet at
; lines
and
meet at
. Show that
.
Solution
Given a triangle,
and a point
in its interior, assume that the circumcircles of
and
are tangent to
. Prove that ray
bisects
.
Let the intersection of
and
be
. By power of a point,
and
, so
.
Let ray
intersect
at
. By our lemma,
,
bisects
. Since
and
are similar, and
and
are similar implies
bisects
.
Now, since
is parallel to
. But
is tangent to the circumcircle of
hence
and that implies
So
is isosceles and
.
By simple parallel line rules, =\angle{ABM}
\angle{BAM}=\angle{EAB}
\textit{ASA}
\triangle{ABM}
\triangle{ABE}$are congruent.
We know that$ (Error compiling LaTeX. Unknown error_msg)BE=BM=BDED
\triangle{EMD}
E
M
EM
PQ
MP = MQ
\triangle{EPQ}
EP = EQ$ .
See Also
2000 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |