Difference between revisions of "2022 USAJMO Problems/Problem 4"
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Let's draw (<math>\ell</math>) perpendicular bisector of <math>\overline{KL}</math>. Let <math>X, Y</math> be intersections of <math>\ell</math> with <math>AC</math> and <math>BD</math>, respectively. <math>KXLY</math> is a kite. Let <math>O</math> mid-point of <math>\overline{KL}</math>. Let <math>M</math> mid-point of <math>\overline{BD}</math> (and also <math>M</math> is mid-point of <math>\overline{AC}</math>). <math>X, O, Y</math> are on the line <math>\ell</math>. | Let's draw (<math>\ell</math>) perpendicular bisector of <math>\overline{KL}</math>. Let <math>X, Y</math> be intersections of <math>\ell</math> with <math>AC</math> and <math>BD</math>, respectively. <math>KXLY</math> is a kite. Let <math>O</math> mid-point of <math>\overline{KL}</math>. Let <math>M</math> mid-point of <math>\overline{BD}</math> (and also <math>M</math> is mid-point of <math>\overline{AC}</math>). <math>X, O, Y</math> are on the line <math>\ell</math>. |
Revision as of 05:49, 15 May 2022
Problem
Let be a rhombus, and let
and
be points such that
lies inside the rhombus,
lies outside the rhombus, and
. Prove that there exist points
and
on lines
and
such that
is also a rhombus.
Solution
(Image of the solution is here [1])
Let's draw () perpendicular bisector of
. Let
be intersections of
with
and
, respectively.
is a kite. Let
mid-point of
. Let
mid-point of
(and also
is mid-point of
).
are on the line
.
,
,
and so
(side-side-side). By spiral similarity,
. Hence, we get
Similarly, ,
,
and so
(side-side-side). From spiral similarity,
. Thus,
If we can show that , then the kite
will be a rhombus.
By spiral similarities, and
. Then,
.
. Then,
. Also, in the right triangles
and
,
. Therefore,
and we get
.
(Lokman GÖKÇE)