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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 and <math>LaTeX</math> by Sp3nc3r
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Solution and <math>\LaTeX</math> by Sp3nc3r
  
 
{{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 15:24, 22 October 2020

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

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