Let ABCD be a convex quadrilateral such that $AB + CD = \sqrt{2}AC$ and $BC + DA = \sqrt{2}BD$. Prove that ABCD is a parallelogram.
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Tags: geometry, parallelogram, inequalities, trigonometry, trig identities, Law of Cosines, geometry unsolved
17.04.2012 13:07
Let $ABEC,AFBC$ be parallelograms. Then $B$ is the midpoint of $EF$. From the conditions we have $\sqrt2 BE=\sqrt2 AC=AB+CD=CE+CD\geq DE$ and $\sqrt2 BD=AD+BC=AD+FA\geq DF$, with both equality iff $ABCD$ is a parallelogram. Now we just consider $DFBE$. Let $F'$ be on the ray $BF$ such that $\angle BDE=\angle DF'E$. Then $EB\cdot EF'=ED^2$, so $EF'\leq EF$. But $\frac{F'D}{DB}=\frac{DE}{EB}=\frac{F'E}{ED}$, so $\left(\frac{F'D}{DB}\right)^2=\frac{EB}{EF'}\leq 2$. This means that equality must hold for both inequality hence $ABCD$ is a parallelogram.
27.04.2012 17:43
oneplusone wrote: Let $ABEC,AFBC$ be parallelograms. Then $B$ is the midpoint of $EF$. From the conditions we have $\sqrt2 BE=\sqrt2 AC=AB+CD=CE+CD\geq DE$ and $\sqrt2 BD=AD+BC=AD+FA\geq DF$, with both equality iff $ABCD$ is a parallelogram. Now we just consider $DFBE$. Let $F'$ be on the ray $BF$ such that $\angle BDE=\angle DF'E$. Then $EB\cdot EF'=ED^2$, so $EF'\leq EF$. But $\frac{F'D}{DB}=\frac{DE}{EB}=\frac{F'E}{ED}$, so $\left(\frac{F'D}{DB}\right)^2=\frac{EB}{EF'}\leq 2$. This means that equality must hold for both inequality hence $ABCD$ is a parallelogram. Great idea of constructing parallelograms $ABEC,AFBC$. By the way, we could also use the Law of Cosines to finish it off: \begin{eqnarray*} DE^2+DF^2 & = & BD^2+BE^2-2 BD \cdot BE \cos(\angle DBE) \\ & + & BD^2+BF^2-2 BD \cdot BF \cos(\pi-\angle DBE) \\ &= & 2BD^2 + 2 AC^2, \end{eqnarray*} which implies that equality must hold for both $\sqrt2 AC\geq DE$ and $\sqrt2 BD\geq DF$, hence $ABCD$ is a parallelogram. [asy][asy] import graph; size(6cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -1.34, xmax = 10.38, ymin = 1.48, ymax = 8.28; draw((3.34,2.62)--(0.66,5.08)); draw((0.66,5.08)--(4.08,4.68)); draw((4.08,4.68)--(6.28,3.42)); draw((6.28,3.42)--(3.34,2.62)); draw((4.08,4.68)--(3.34,2.62)); draw((0.66,5.08)--(6.28,3.42)); draw((1.4,7.14)--(4.08,4.68)); draw((0.66,5.08)--(-0.08,3.02)); draw((-0.08,3.02)--(3.34,2.62)); draw((1.4,7.14)--(0.66,5.08)); draw((1.4,7.14)--(6.28,3.42), red); draw((-0.08,3.02)--(6.28,3.42), red); dot((3.34,2.62),dotstyle); label("$A$", (3.48,2.34), NE * labelscalefactor); dot((0.66,5.08),dotstyle); label("$B$", (0.4-0.2,5.1), NE * labelscalefactor); dot((4.08,4.68),dotstyle); label("$C$", (4.02,4.26-0.2), NE * labelscalefactor); dot((6.28,3.42),dotstyle); label("$D$", (6.36,3.54), NE * labelscalefactor); dot((1.4,7.14),dotstyle); label("$E$", (1.48,7.26), NE * labelscalefactor); dot((-0.08,3.02),dotstyle); label("$F$", (-0.4-0.2,2.86), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]