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= | + | =Solution 2 (similar to Solution 1)= |
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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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>. | 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> | + | 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 22:58, 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.