Difference between revisions of "2020 USAMTS Round 1 Problems/Problem 3"
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By the same logic, <math>HF \parallel EG \Rightarrow GFHE</math> is a parallelogram. | By the same logic, <math>HF \parallel EG \Rightarrow GFHE</math> is a parallelogram. | ||
2. <math>\angle EAB = \frac{\angle DAB}{2}</math> and <math>\angle ABE = \frac{\angle ABC}{2} \Rightarrow \angle EAB + \angle ABE = \frac{\angle DAB + \angle ABC}{2}</math> and <math>\angle DAB + \angle ABC = 180^\circ \Rightarrow \angle EAB + \angle ABE = 90^\circ \Rightarrow \angle AEB = 90^\circ.</math> | 2. <math>\angle EAB = \frac{\angle DAB}{2}</math> and <math>\angle ABE = \frac{\angle ABC}{2} \Rightarrow \angle EAB + \angle ABE = \frac{\angle DAB + \angle ABC}{2}</math> and <math>\angle DAB + \angle ABC = 180^\circ \Rightarrow \angle EAB + \angle ABE = 90^\circ \Rightarrow \angle AEB = 90^\circ.</math> | ||
− | By <math>1</math> and <math>2,</math> we can conclude that <math>HFGE</math> is a rectangle. <math> | + | By <math>1</math> and <math>2,</math> we can conclude that <math>HFGE</math> is a rectangle. |
+ | |||
+ | Now, knowing <math>HFGE</math> is a rectangle, we can continue on. | ||
+ | |||
Let <math>AB = a, BC = b, </math> and <math>\angle ABE = \alpha.</math> Thus, <math>[ABCD] = ab\sin(2\alpha).</math> | Let <math>AB = a, BC = b, </math> and <math>\angle ABE = \alpha.</math> Thus, <math>[ABCD] = ab\sin(2\alpha).</math> | ||
<math>AD \parallel DC \Rightarrow \angle BJC = \angle JCD</math> and <math>\angle JCD = \angle JCB \Rightarrow \angle BJC = \angle JCB \Rightarrow JB = BC =b.</math> | <math>AD \parallel DC \Rightarrow \angle BJC = \angle JCD</math> and <math>\angle JCD = \angle JCB \Rightarrow \angle BJC = \angle JCB \Rightarrow JB = BC =b.</math> | ||
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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 |
+ | |||
+ | =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> \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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | {{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.