sqrt([AKON]) + sqrt([CLOM]) <= sqrt([ABCD])

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sqrt([AKON]) + sqrt([CLOM]) <= sqrt([ABCD])

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Source: IMO Shortlist 1995, G7

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nttu

#1 h Dec 30, 2005, 10:28 AM • 2 Y

Y by Adventure10, Mango247

Let ABCD be a convex quadrilateral and O a point inside it. Let the parallels to the lines BC, AB, DA, CD through the point O meet the sides AB, BC, CD, DA of the quadrilateral ABCD at the points E, F, G, H, respectively. Then, prove that $\sqrt {\left|AHOE\right|} + \sqrt {\left|CFOG\right|}\leq\sqrt {\left|ABCD\right|}$ , where $\left|P_1P_2...P_n\right|$ is an abbreviation for the non-directed area of an arbitrary polygon $P_1P_2...P_n$ .

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#2 h Feb 25, 2006, 12:30 AM • 2 Y

Y by Adventure10, Mango247

Problem. Let ABCD be a convex quadrilateral and O a point inside it. Let the parallels to the lines BC, AB, DA, CD through the point O meet the sides AB, BC, CD, DA of the quadrilateral ABCD at the points E, F, G, H, respectively. Then, prove that $\sqrt {\left|AHOE\right|} + \sqrt {\left|CFOG\right|}\leq\sqrt {\left|ABCD\right|}$ , where $\left|P_1P_2...P_n\right|$ is an abbreviation for the non-directed area of an arbitrary polygon $P_1P_2...P_n$ .

This problem is, indeed, problem G7 from the IMO Shortlist 1995, and respawned as problem 3 in the 2nd round of the German competition BWM (Bundeswettbewerb Mathematik) 1999.

Solution. Let g be the parallel to the line BD through the point O. Let this line g meet the segment AC at a point O'. The problem will be solved once we show that $\sqrt {\left|AHOE\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ and $\sqrt {\left|CFOG\right|}\leq\frac {CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ ; in fact, these relations will yield

$\sqrt {\left|AHOE\right|} + \sqrt {\left|CFOG\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|} + \frac {CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$
$= \frac {AO^{\prime} + CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|} = \frac {AC}{AC}\cdot\sqrt {\left|ABCD\right|} = \sqrt {\left|ABCD\right|}$ ,

and this is what we have to prove.

We will only show that $\sqrt {\left|AHOE\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ ; the relation $\sqrt {\left|CFOG\right|}\leq\frac {CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ will then follow by analogy.

Let the parallels to the lines BC and CD through the point O' meet the lines AB and DA at the points E' and H', respectively. Then, by Thales, $\frac {AH^{\prime}}{AD} = \frac {AO^{\prime}}{AC}$ and $\frac {AE^{\prime}}{AB} = \frac {AO^{\prime}}{AC}$ , so that the homothety with center A and factor $\frac {AO^{\prime}}{AC}$ maps the points C, D and B to the points O', H' and E', respectively. This has two important consequences: first, E'H' || BD (since homotheties map lines to parallel lines), and then, $\frac {\left|AE^{\prime}O^{\prime}H^{\prime}\right|}{\left|ABCD\right|} = \left(\frac {AO^{\prime}}{AC}\right)^2$ (since homotheties multiply areas with the squared homothety factor). The latter equation yields $\left|AE^{\prime}O^{\prime}H^{\prime}\right| = \left(\frac {AO^{\prime}}{AC}\right)^2\cdot\left|ABCD\right|$ , so that $\sqrt {\left|AE^{\prime}O^{\prime}H^{\prime}\right|} = \frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ ; therefore, in order to prove $\sqrt {\left|AHOE\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ , it is enough to show that $\left|AHOE\right|\leq\left|AE^{\prime}O^{\prime}H^{\prime}\right|$ .

Let the line g meet the lines AB and DA at the points U and V, respectively. Since $\left|AHOE\right| = \left|AUV\right| - \left(\left|UOE\right| + \left|VOH\right|\right)$ and $\left|AE^{\prime}O^{\prime}H^{\prime}\right| = \left|AUV\right| - \left(\left|UO^{\prime}E^{\prime}\right| + \left|VO^{\prime}H^{\prime}\right|\right)$ , proving that $\left|AHOE\right|\leq\left|AE^{\prime}O^{\prime}H^{\prime}\right|$ is equivalent to proving that $\left|UOE\right| + \left|VOH\right|\geq\left|UO^{\prime}E^{\prime}\right| + \left|VO^{\prime}H^{\prime}\right|$ . This can be done as follows:

Let e be the distance from the point E to the line g, or, equivalently, the length of the E-altitude in triangle UOE. Then, the area of triangle UOE equals $\left|UOE\right| = \frac12\cdot UO\cdot e$ (in fact, the area of a triangle equals $\frac12\cdot$ sidelength $\cdot$ corresponding altitude). Similarly, if h is the distance from the point H to the line g, then $\left|VOH\right| = \frac12\cdot VO\cdot h$ . Thus,

$\left|UOE\right| + \left|VOH\right| = \frac12\cdot UO\cdot e + \frac12\cdot VO\cdot h = \frac12\cdot\left(UO\cdot e + VO\cdot h\right)$ .

Similarly, working with the point O' instead of O, we get

$\left|UO^{\prime}E^{\prime}\right| + \left|VO^{\prime}H^{\prime}\right| = \frac12\cdot\left(UO^{\prime}\cdot e^{\prime} + VO^{\prime}\cdot h^{\prime}\right)$ ,

where e' and h' are the distances from the points E' and H' to the line g, respectively. Thus, in order to prove $\left|UOE\right| + \left|VOH\right|\geq\left|UO^{\prime}E^{\prime}\right| + \left|VO^{\prime}H^{\prime}\right|$ , it is enough to show that

$UO\cdot e + VO\cdot h\geq UO^{\prime}\cdot e^{\prime} + VO^{\prime}\cdot h^{\prime}$ .

Since OE || BC and O'E' || BC, we have OE || O'E', and thus, there is a homothety with center U which maps the points O and E to the points O' and E'. This homothety must obviously leave the point U fixed (since U is the center of this homothety). Thus, the triangles UOE and UO'E' are similar. In similar triangles, the ratio of a sidelength to the corresponding altitude is equal; thus, in the triangles UOE and UO'E', we have $\frac {UO}{e} = \frac {UO^{\prime}}{e^{\prime}}$ (in fact, e and e' are the altitudes of these triangles corresponding to the sidelengths UO and UO'). Similarly, $\frac {VO}{h} = \frac {VO^{\prime}}{h^{\prime}}$ .

Now, the points E' and H' have a property which the points E and H don't have (at least not necessarily): Since E'H' || BD and g || BD, we have E'H' || g, so that the distances e' and h' from the points E' and H' to the line g are equal. Thus,

$UO^{\prime}\cdot e^{\prime} + VO^{\prime}\cdot h^{\prime} = UO^{\prime}\cdot e^{\prime} + VO^{\prime}\cdot e^{\prime}$
$= \left(UO^{\prime} + VO^{\prime}\right)\cdot e^{\prime} = UV\cdot e^{\prime} = \frac {UV^2}{\left(\frac {UV}{e^{\prime}}\right)} = \frac {\left(UO + VO\right)^2}{\left(\frac {UO^{\prime} + VO^{\prime}}{e^{\prime}}\right)}$
$= \frac {\left(UO + VO\right)^2}{\frac {UO^{\prime}}{e^{\prime}} + \frac {VO^{\prime}}{e^{\prime}}} = \frac {\left(UO + VO\right)^2}{\frac {UO^{\prime}}{e^{\prime}} + \frac {VO^{\prime}}{h^{\prime}}} = \frac {\left(UO + VO\right)^2}{\frac {UO}{e} + \frac {VO}{h}}$ .

So the inequality in question, $UO\cdot e + VO\cdot h\geq UO^{\prime}\cdot e^{\prime} + VO^{\prime}\cdot h^{\prime}$ , becomes $UO\cdot e + VO\cdot h\geq\frac {\left(UO + VO\right)^2}{\frac {UO}{e} + \frac {VO}{h}}$ , or, equivalently, $\left(UO\cdot e + VO\cdot h\right)\left(\frac {UO}{e} + \frac {VO}{h}\right)\geq\left(UO + VO\right)^2$ , what is trivial from Cauchy-Schwarz:

$\left(UO\cdot e + VO\cdot h\right)\left(\frac {UO}{e} + \frac {VO}{h}\right)\geq\left(\sqrt {\left(UO\cdot e\right)\cdot\frac {UO}{e}} + \sqrt {\left(VO\cdot h\right)\cdot\frac {VO}{h}}\right)^2$
$= \left(UO + VO\right)^2$ .

This completes the solution of the problem.

Darij

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Wizzy

#3 h Aug 16, 2008, 7:40 AM • 4 Y

Y by Adventure10, Mango247, popop614, kiyoras_2001

This problem is simply following from a Brunn-Minkovsky inequality. Am I right?

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