Difference between revisions of "2020 USOJMO Problems/Problem 4"
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Let <math>G</math> be the intersection of <math>AE</math> and <math>(ABCD)</math> and <math>H</math> be the intersection of <math>DF</math> and <math>(ABCD)</math>. | Let <math>G</math> be the intersection of <math>AE</math> and <math>(ABCD)</math> and <math>H</math> be the intersection of <math>DF</math> and <math>(ABCD)</math>. | ||
− | [b][color= | + | [b][color=#f00]Claim: <math>GH || FE || BC</math>[/color][/b] |
By Pascal's on <math>GDCBAH</math>, we see that the intersection of <math>GH</math> and <math>BC</math>, <math>E</math>, and <math>F</math> are collinear. Since <math>FE || BC</math>, we know that <math>HG || BC</math> as well. <math>\blacksquare</math> | By Pascal's on <math>GDCBAH</math>, we see that the intersection of <math>GH</math> and <math>BC</math>, <math>E</math>, and <math>F</math> are collinear. Since <math>FE || BC</math>, we know that <math>HG || BC</math> as well. <math>\blacksquare</math> | ||
− | [b][color= | + | [b][color=#f00]Claim: <math>FB = FD</math>[/color][/b] |
Note that since all cyclic trapezoids are isosceles, <math>HB = GC</math>. Since <math>AB = BC</math> and <math>EB \perp AC</math>, we know that <math>EA = EC</math>, from which we have that <math>DGCA</math> is an isosceles trapezoid and <math>DA = GC</math>. It follows that <math>DA = GC = HB</math>, so <math>BHAD</math> is an isosceles trapezoid, from which <math>FB = FD</math>, as desired. <math>\blacksquare</math> | Note that since all cyclic trapezoids are isosceles, <math>HB = GC</math>. Since <math>AB = BC</math> and <math>EB \perp AC</math>, we know that <math>EA = EC</math>, from which we have that <math>DGCA</math> is an isosceles trapezoid and <math>DA = GC</math>. It follows that <math>DA = GC = HB</math>, so <math>BHAD</math> is an isosceles trapezoid, from which <math>FB = FD</math>, as desired. <math>\blacksquare</math> |
Revision as of 23:48, 22 June 2020
Let be the intersection of
and
and
be the intersection of
and
.
[b][color=#f00]Claim:
[/color][/b]
By Pascal's on
, we see that the intersection of
and
,
, and
are collinear. Since
, we know that
as well.
[b][color=#f00]Claim:
[/color][/b]
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.