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

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 $AB > BC$ , $r = \boxed{ 2+ \sqrt{3}}$ . Template:USAMTS box The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2020 USAMTS Round 1 Problems/Problem 3

Solution 1

Solution 2