Difference between revisions of "2019 USAMO Problems/Problem 2"
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+ | Realize that there is only one point <math>P</math> on <math>\overline{AB}</math> satisfying the conditions, because <math>\angle APD</math> decreases and <math>\angle BPC</math> increases as <math>P</math> moves from <math>A</math> to <math>B</math>. Therefore, if we prove that there is a single point <math>P</math> that lies on <math>\overline{AB}</math> such that <math>\angle APD \cong \angle BPC</math>, and that <math>PE</math> bisects <math>CD</math>, it must coincide with the point from the problem, so we will be done. | ||
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+ | Since <math>AD^2 + BC^2 = AB^2</math>, there is some <math>P</math> on <math>AB</math> such that <cmath>AD^2 = AP \times AB \text{ and } BC^2 = BP \times BA.</cmath> Thus, <math>\frac{AP}{AD} = \frac{AD}{AB}</math> and <math>\frac{BP}{BC} = \frac{BC}{BA}</math>. Thus we have that <math>\triangle APD \sim | ||
+ | \triangle ADB</math> and <math>\triangle BPC \sim \triangle BCA</math>, meaning that <math>\angle APD = \angle ADB = \angle ACB = \angle BPC</math>. | ||
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+ | We next must show that <math>PE</math> bisects <math>CD</math>. Define <math>K</math> as the intersection of <math>AC</math> and <math>PD</math> and <math>L</math> as the intersection of <math>BD</math> and <math>PC</math>. We know that <math>APLD</math> and <math>BPKC</math> are cyclic, because | ||
+ | <cmath>\angle'ADL = \angle'ACB = \angle'BPC = \angle'AP L,</cmath> where <math>\angle'</math> represents an angle which is measured <math>\text{mod } \pi</math>. Furthermore, quadrilateral <math>AKLB</math> is also cyclic, because | ||
+ | <cmath>\angle'AKB = \angle'CKB = \angle'CP B</cmath> and <math>\angle'ALB = \angle'APD</math>, and these are equal. | ||
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+ | As the quadrilaterals are cyclic, we have that <math>\angle'KCD = \angle'ABD = \angle'ABL = \angle'AKL = \angle'CKL</math>, meaning that <math>CD \parallel KL</math>. Thus <math>CDKL</math> is a trapezoid whose legs intersect at <math>P</math> and whose diagonals intersect at <math>E</math>. Therefore, line <math>PE</math> bisects the bases <math>CD</math> and <math>KL</math>, as wished. ~ciceronii | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:03, 27 February 2020
Problem
Let be a cyclic quadrilateral satisfying
. The diagonals of
intersect at
. Let
be a point on side
satisfying
. Show that line
bisects
.
Solution
Realize that there is only one point on
satisfying the conditions, because
decreases and
increases as
moves from
to
. Therefore, if we prove that there is a single point
that lies on
such that
, and that
bisects
, it must coincide with the point from the problem, so we will be done.
Since , there is some
on
such that
Thus,
and
. Thus we have that
and
, meaning that
.
We next must show that bisects
. Define
as the intersection of
and
and
as the intersection of
and
. We know that
and
are cyclic, because
where
represents an angle which is measured
. Furthermore, quadrilateral
is also cyclic, because
and
, and these are equal.
As the quadrilaterals are cyclic, we have that , meaning that
. Thus
is a trapezoid whose legs intersect at
and whose diagonals intersect at
. Therefore, line
bisects the bases
and
, as wished. ~ciceronii
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |