Difference between revisions of "2020 USOJMO Problems/Problem 4"
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\angle{FEB}</math>, leads to <math>FE=FB</math> | \angle{FEB}</math>, leads to <math>FE=FB</math> | ||
− | Notice that quadrilateral <math>AFED</math> is a cyclic quadrilateral since <math>\angle{ADE}+\angle{AFE}=\angle{ADE}+\angle{ABC}=180^{\circ}. Now notice that </math>\ | + | Notice that quadrilateral <math>AFED</math> is a cyclic quadrilateral since <math>\angle{ADE}+\angle{AFE}=\angle{ADE}+\angle{ABC}=180^{\circ}. Now notice that </math>\widehat{SE}=\widehat{SE}, \angle{SAE}=\angle{SDE}=\beta<math>; similarly, </math>\widehat{FS}=\widehat{FS}; <math>\angle{FDS}=\angle{FAS}=\alpha</math> |
<math>\angle{FDE}=\alpha+\beta; \angle{FED}=\angle{BCD}=\alpha+\beta</math>, it leads to <math>FD=FE</math>. | <math>\angle{FDE}=\alpha+\beta; \angle{FED}=\angle{BCD}=\alpha+\beta</math>, it leads to <math>FD=FE</math>. | ||
since <math>FE=FB;FD=FE, DF=BF</math> as desired | since <math>FE=FB;FD=FE, DF=BF</math> as desired | ||
~bluesoul | ~bluesoul |
Revision as of 05:28, 9 January 2022
Contents
Problem
Let be a convex quadrilateral inscribed in a circle and satisfying
. Points
and
are chosen on sides
and
such that
and
. Prove that
.
Solution
Let be the intersection of
and
and
be the intersection of
and
.
Claim:
By Pascal's on , we see that the intersection of
and
,
, and
are collinear. Since
, we know that
as well.
Note that since all cyclic trapezoids are isosceles, . Since
and
, we know that
, from which we have that
is an isosceles trapezoid and
. It follows that
, so
is an isosceles trapezoid, from which
, as desired.
Solution 2
Let , and let
. Now let
and
.
From and
, we have
so
. From cyclic quadrilateral ABCD,
. Since
,
.
Now from cyclic quadrilateral ABC and we have
. Thus F, A, D, and E are concyclic, and
Let this be statement 1.
Now since , triangle ABC gives us
. Thus
, or
.
Right triangle BHC gives , and
implies
Now triangle BGE gives . But
, so
. Using triangle FGD and statement 1 gives
Thus, , so
as desired.
~MortemEtInteritum
Solution 3 (Angle-Chasing)
Proving that is equivalent to proving that
. Note that
because quadrilateral
is cyclic. Also note that
because
.
, which follows from the facts that
and
, implies that
. Thus, we would like to prove that triangle
is similar to triangle
. In order for this to be true, then
must equal
which implies that
must equal
. In order for this to be true, then quadrilateral
must be cyclic. Using the fact that
, we get that
, and that
, and thus we have proved that quadrilateral
is cyclic. Therefore, triangle
is similar to isosceles triangle
from AA and thus
.
-xXINs1c1veXx
Solution 4
BE is perpendicular bisector of AC, so . FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic.
. Hence,
,
.
Mathdummy
Solution 5
Let be on
such that
, and
. Then
is the orthic triangle of
. Thus,
is the midpoint of
and lies on the
bisector of
.
Solution 6
Let meet
at
,
meet
at
, connect
.
Denote that
, since
is parallel to
,
.
and
are vertical angle, so they are equal to each other.
,
, since
, we can express
, leads to
Notice that quadrilateral is a cyclic quadrilateral since
\widehat{SE}=\widehat{SE}, \angle{SAE}=\angle{SDE}=\beta
\widehat{FS}=\widehat{FS};
, it leads to
.
since
as desired
~bluesoul