View source for 2020 USAMTS Round 1 Problems/Problem 3

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

=Solution 1=
We claim the answer is <math>2+\sqrt3.</math> Let <math>HFGE</math> be the new quadrilateral; that is, the quadrilateral determined by the internal bisectors of the angles of <math>ABCD</math>.

Lemma <math>1</math> : <math>HFGE</math> is a rectangle.
<math>1.</math> <math>ABCD</math> is a parallelogram.
<math>\angle DAB = \angle DCB,</math> as <math>AE</math> bisects <math>\angle DAB \Rightarrow \angle BAE = \frac{\angle DAB}{2}</math> and <math>CE</math> bisects <math>\angle DCB \Rightarrow \angle DCF = \frac{DCB}{2} \Rightarrow \angle DCF = \angle AJF \Rightarrow \angle BAE = \angle AJF \Rightarrow FG \parallel HE.</math>
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>
By <math>1</math> and <math>2,</math> we can conclude that <math>HFGE</math> is a rectangle. <math>\blacksquare</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>
By the same logic, <math>AI = AD = b.</math>
<math>BE \parallel ED \Rightarrow \angle AIH = \angle ABE = \alpha.</math>
<math>HE = AE-AH = a\sin(\alpha) - b\sin(\alpha) = (\alpha - \beta)\sin(\alpha),</math> and <math>EG = EB-GB = a\cos(\alpha) - b\cos(\alpha) = (a-b)\cos(\alpha).</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