Difference between revisions of "2020 USAMTS Round 1 Problems/Problem 3"
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Solution and <math>\LaTeX</math> by Sp3nc3r | Solution and <math>\LaTeX</math> by Sp3nc3r | ||
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Let <math>P,Q,R,S</math> be the intersections of the bisectors of <math>\angle C \text { and } \angle D, \angle B \text { and } \angle C, \angle A \text { and } \angle B, \angle A \text { and } \angle D</math> respectively. | Let <math>P,Q,R,S</math> be the intersections of the bisectors of <math>\angle C \text { and } \angle D, \angle B \text { and } \angle C, \angle A \text { and } \angle B, \angle A \text { and } \angle D</math> respectively. | ||
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Therefore, by the Quadratic Formula, <math>r= 2 \pm \sqrt{3}</math>. Since <math> AB > BC</math>, <math>r = \boxed{ 2+ \sqrt{3}}</math>. | Therefore, by the Quadratic Formula, <math>r= 2 \pm \sqrt{3}</math>. Since <math> AB > BC</math>, <math>r = \boxed{ 2+ \sqrt{3}}</math>. | ||
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{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:58, 3 June 2022
The bisectors of the internal angles of parallelogram with
determine a quadrilateral with the same area as
. Determine, with proof, the value of
.
Solution 1
We claim the answer is Let
be the new quadrilateral; that is, the quadrilateral determined by the internal bisectors of the angles of
.
Lemma :
is a rectangle.
is a parallelogram.
as
bisects
and
bisects
By the same logic,
is a parallelogram.
2.
and
and
By
and
we can conclude that
is a rectangle.
Now, knowing is a rectangle, we can continue on.
Let and
Thus,
and
By the same logic,
and
Because
we have
Solution and by Sp3nc3r
Solution 2 (similar to Solution 1)
Let be the intersections of the bisectors of
respectively.
Let . Then
and
. So,
. Therefore,
.
Similarly, .
So, therefore, must be a rectangle and
Now, note that . Also,
.
So, we have
Since :
for
.
Therefore, by the Quadratic Formula, . Since
,
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.