Difference between revisions of "2020 USAMTS Round 1 Problems/Problem 3"
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=Solution 2= | =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. | + | 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> \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>. | + | 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>. |
− | Similarly, <math>\angle SPQ = \angle QRS = 180- (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90</math>. | + | Similarly, <math>\angle SPQ = \angle QRS = 180- (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90</math>. |
− | So, therefore, <math>PQRS</math> must be a rectangle and <math>[PQRS] = SP \times RS</math> | + | So, therefore, <math>PQRS</math> must be a rectangle and <math>[PQRS] = SP \times RS</math> |
− | 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>. | + | 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>. |
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> | 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> |
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.
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
,
$\qed$ (Error compiling LaTeX. Unknown error_msg)
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