Difference between revisions of "2000 IMO Problems/Problem 1"
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− | <math>\textbf{Lemma:}</math> Given a triangle, <math>ABC</math> and a point <math>P</math> in | + | <math>\textbf{Lemma:}</math> Given a triangle, <math>ABC</math> and a point <math>P</math> in its interior, assume that the circumcircles of <math>\triangle{ACP}</math> and <math>\triangle{ABP}</math> are tangent to <math>BC</math>. Prove that ray <math>AP</math> bisects <math>BC</math>. |
<math>\textbf{Proof:}</math> Let the intersection of <math>AP</math> and <math>BC</math> be <math>D</math>. By power of a point, <math>BD^2=AD(PD)</math> and <math>CD^2=AD(PD)</math>, so <math>BD=CD</math>. | <math>\textbf{Proof:}</math> Let the intersection of <math>AP</math> and <math>BC</math> be <math>D</math>. By power of a point, <math>BD^2=AD(PD)</math> and <math>CD^2=AD(PD)</math>, so <math>BD=CD</math>. | ||
Revision as of 14:05, 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
.
By simple parallel line rules, . Similarly,
, so by
criterion,
and
are congruent.
Now, and
since
is parallel to
. But
is tangent to the circumcircle of
hence
and that implies
So
is isosceles and
.
We know that so a circle with diameter
can be circumscribed around
. Join points
and
,
is perpendicular on
, previously we proved
, hence
is isoscles and
.
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 |