Difference between revisions of "2020 USAMTS Round 1 Problems/Problem 3"
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<math>[HFGE] = HE * EG = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\text{sin}(\alpha)\text{cos}(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow 2ab = (a-b)^2 \Rightarrow a^2 + b^2 -2ab = 2ab</math> <math>\Rightarrow a^2 -4ab +b^2 = 0 \Rightarrow a = \frac{4b \pm \sqrt{16b^2 -4b^2}}{2} = 2b\pm b\sqrt{3}</math> <math>\Rightarrow a=b(2\pm\sqrt{3}) \Rightarrow \frac{a}{b} = 2 \pm \sqrt{3}.</math> Because <math>a>b,</math> we have <math>\frac{a}{b} = 2+\sqrt{3}.</math> | <math>[HFGE] = HE * EG = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\text{sin}(\alpha)\text{cos}(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow 2ab = (a-b)^2 \Rightarrow a^2 + b^2 -2ab = 2ab</math> <math>\Rightarrow a^2 -4ab +b^2 = 0 \Rightarrow a = \frac{4b \pm \sqrt{16b^2 -4b^2}}{2} = 2b\pm b\sqrt{3}</math> <math>\Rightarrow a=b(2\pm\sqrt{3}) \Rightarrow \frac{a}{b} = 2 \pm \sqrt{3}.</math> Because <math>a>b,</math> we have <math>\frac{a}{b} = 2+\sqrt{3}.</math> | ||
− | Solution by Sp3nc3r | + | Solution and <math>LaTeX</math> by Sp3nc3r |
+ | |||
+ | {{USAMTS box|year=2020|round=1|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Revision as of 16:24, 22 October 2020
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
Template:USAMTS box
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.