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 | ||
+ | |||
+ | =Solution 2= | ||
+ | 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. \bigskip | ||
+ | |||
+ | Let <math> \angle BAD = \theta</math> . Then <math>\angle SAD = \angle QCB= \frac{\theta}{2}</math> and <math>\angle ADS = \angle QBC= \frac{180-\theta}{2}</math>. So, <math>\angle ASD = \angle SRQ = \angle PQR = \angle 180 - (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90</math>. Therefore, <math>RSP = 90</math>.\bigskip | ||
+ | |||
+ | Similarly, <math>\angle SPQ = \angle QRS = 180- (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90</math>.\bigskip | ||
+ | |||
+ | So, therefore, <math>PQRS</math> must be a rectangle and <math>[PQRS] = SP \times RS</math> \bigskip | ||
+ | |||
+ | Now, note that <math>SP= PD- SD = DC \sin(\frac{\theta}{2}) - AD \sin(\frac{\theta}{2})</math>. Also, <math>RS = AR - AS - DC \cos(\frac{\theta}{2}) - AD \cos(\frac{\theta}{2})</math>.\bigskip | ||
+ | |||
+ | So, we have <cmath> [PQRS] = (DC-AD)^2 \sin(\frac{\theta}{2}) \cos (\frac{\theta}{2})</cmath> <cmath>[ABCD] = DC \times AD \sin{\theta} = DC \times AD \times 2 \sin(\frac{\theta}{2}) \cos (\frac{\theta}{2}). </cmath> | ||
+ | |||
+ | Since <math>[PQRS] = [ABCD]</math> :<cmath>(DC- AD)^2 = 2(DC)(AD) \implies r^2 - 4r + 1 = 0</cmath> for <math>r = \frac{DC}{AD}</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> <math>\qed</math> | ||
{{USAMTS box|year=2020|round=1|num-b=2|num-a=4}} | {{USAMTS box|year=2020|round=1|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:22, 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
Solution 2
Let be the intersections of the bisectors of
respectively. \bigskip
Let . Then
and
. So,
. Therefore,
.\bigskip
Similarly, .\bigskip
So, therefore, must be a rectangle and
\bigskip
Now, note that . Also,
.\bigskip
So, we have
Since :
for
.
Therefore, by the Quadratic Formula, . Since
,
$\qed$ (Error compiling LaTeX. Unknown error_msg)
Template:USAMTS box
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