Difference between revisions of "2022 USAJMO Problems/Problem 4"
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<cmath> \angle XBD = \angle XDB = \angle XKL = \angle XLK = b .</cmath> | <cmath> \angle XBD = \angle XDB = \angle XKL = \angle XLK = b .</cmath> | ||
− | Similarly, <math>AK =CL</math>, <math>YK = YL</math>, <math>YA=YC</math> and so <math>\triangle | + | Similarly, <math>AK =CL</math>, <math>YK = YL</math>, <math>YA=YC</math> and so <math>\triangle AKY \cong \triangle CYL</math> (side-side-side). From spiral similarity, <math>\triangle YKL\sim \triangle YAC</math>. Thus, |
<cmath> \angle YAC = \angle YCA = \angle YKL = \angle YLK = a .</cmath> | <cmath> \angle YAC = \angle YCA = \angle YKL = \angle YLK = a .</cmath> | ||
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(Lokman GÖKÇE) | (Lokman GÖKÇE) | ||
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+ | ==See Also== | ||
+ | {{USAJMO newbox|year=2022|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 08:46, 13 February 2024
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)
See Also
2022 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.