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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>
  
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Solution by Sp3nc3r
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Solution and <math>\LaTeX</math> by Sp3nc3r
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=Solution 2 (similar to Solution 1)=
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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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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>.
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Similarly, <math>\angle SPQ = \angle QRS = 180-  (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90</math>.
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So, therefore, <math>PQRS</math> must be a rectangle and <math>[PQRS] = SP \times RS</math>
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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>.
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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>
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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>.
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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}}

Latest revision as of 22:58, 3 June 2022

The bisectors of the internal angles of parallelogram $ABCD$ with $AB>BC$ determine a quadrilateral with the same area as $ABCD$. Determine, with proof, the value of $\frac{AB}{BC}$.

Solution 1

We claim the answer is $2+\sqrt3.$ Let $HFGE$ be the new quadrilateral; that is, the quadrilateral determined by the internal bisectors of the angles of $ABCD$.

Lemma $1$ : $HFGE$ is a rectangle. $1.$ $ABCD$ is a parallelogram. $\angle DAB = \angle DCB,$ as $AE$ bisects $\angle DAB \Rightarrow \angle BAE = \frac{\angle DAB}{2}$ and $CE$ bisects $\angle DCB \Rightarrow \angle DCF = \frac{DCB}{2} \Rightarrow \angle DCF = \angle AJF \Rightarrow \angle BAE = \angle AJF \Rightarrow FG \parallel HE.$ By the same logic, $HF \parallel EG \Rightarrow GFHE$ is a parallelogram. 2. $\angle EAB = \frac{\angle DAB}{2}$ and $\angle ABE = \frac{\angle ABC}{2} \Rightarrow \angle EAB + \angle ABE = \frac{\angle DAB + \angle ABC}{2}$ and $\angle DAB + \angle ABC = 180^\circ \Rightarrow \angle EAB + \angle ABE = 90^\circ \Rightarrow \angle AEB = 90^\circ.$ By $1$ and $2,$ we can conclude that $HFGE$ is a rectangle.

Now, knowing $HFGE$ is a rectangle, we can continue on.

Let $AB = a, BC = b,$ and $\angle ABE = \alpha.$ Thus, $[ABCD] = ab\sin(2\alpha).$ $AD \parallel DC \Rightarrow \angle BJC = \angle JCD$ and $\angle JCD = \angle JCB \Rightarrow \angle BJC = \angle JCB \Rightarrow JB = BC =b.$ By the same logic, $AI = AD = b.$ $BE \parallel ED \Rightarrow \angle AIH = \angle ABE = \alpha.$ $HE = AE-AH = a\sin(\alpha) - b\sin(\alpha) = (\alpha - \beta)\sin(\alpha),$ and $EG = EB-GB = a\cos(\alpha) - b\cos(\alpha) = (a-b)\cos(\alpha).$ $[HFGE] = HE * EG = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\text{sin}(\alpha)\text{cos}(\alpha) \Rightarrow ab\sin(2\alpha) = (a-b)^2\sin(\alpha)\cos(\alpha) \Rightarrow 2ab = (a-b)^2 \Rightarrow a^2 + b^2 -2ab = 2ab$ $\Rightarrow a^2 -4ab +b^2 = 0 \Rightarrow a = \frac{4b \pm \sqrt{16b^2 -4b^2}}{2} = 2b\pm b\sqrt{3}$ $\Rightarrow a=b(2\pm\sqrt{3}) \Rightarrow \frac{a}{b} = 2 \pm \sqrt{3}.$ Because $a>b,$ we have $\frac{a}{b} = 2+\sqrt{3}.$

Solution and $\LaTeX$ by Sp3nc3r

Solution 2 (similar to Solution 1)

Let $P,Q,R,S$ be the intersections of the bisectors of $\angle C \text { and } \angle D, \angle B \text { and } \angle C, \angle A \text { and } \angle B, \angle A \text { and } \angle D$ respectively.

Let $\angle BAD = \theta$ . Then $\angle SAD = \angle QCB= \frac{\theta}{2}$ and $\angle ADS = \angle QBC= \frac{180-\theta}{2}$. So, $\angle ASD = \angle SRQ = \angle PQR = \angle 180 - (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90$. Therefore, $RSP = 90$.

Similarly, $\angle SPQ = \angle QRS = 180- (\frac{\theta}{2} + \frac{180-\theta}{2}) = 90$.

So, therefore, $PQRS$ must be a rectangle and $[PQRS] = SP \times RS$

Now, note that $SP= PD- SD = DC \sin(\frac{\theta}{2}) - AD \sin(\frac{\theta}{2})$. Also, $RS = AR - AS - DC \cos(\frac{\theta}{2}) - AD \cos(\frac{\theta}{2})$.

So, we have \[[PQRS] = (DC-AD)^2 \sin(\frac{\theta}{2}) \cos (\frac{\theta}{2})\] \[[ABCD] = DC \times AD \sin{\theta} = DC \times AD \times 2 \sin(\frac{\theta}{2}) \cos (\frac{\theta}{2}).\]

Since $[PQRS] = [ABCD]$ :\[(DC- AD)^2 = 2(DC)(AD) \implies r^2 - 4r + 1 = 0\] for $r = \frac{DC}{AD}$.

Therefore, by the Quadratic Formula, $r= 2 \pm \sqrt{3}$. Since $AB > BC$, $r = \boxed{ 2+ \sqrt{3}}$.


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