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lemma28

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I'm a collage teacher and I've found a very hard problem in one of my math classrooms' textbooks. It was firstly proposed as problem n. 9, back in 1995, in the "

The text is rather simple:

Practically there's a "deltoid" (a quadrilateral with perpendicular diagonals)

Till now I've found that:

Since I'm not satisfied with the analytical cartesian way (with huge calculations) and I don't think that was the proof intended by the people that proposed this problem back in 1995 and since I think there must be a clever and a more elegant geometrical way to it, I'm asking for

I've also challenged my math students (aged 16-18, about 70 students) to find a solution to this problem, and I'm curious to see if a fresher (and just less experienced) brain than mine can find the right way when my

[itex]\frac{1}{\bar{AP} \cdot \bar{AP'}}= \frac{b^2+a^2 m^2}{\left(a^2b^2-b^2 x_A{}^2-a^2y_A{}^2\right) \left(1+m^2\right)} [/itex]

and

[itex]\frac{1}{\text{AQ} \text{AQ}'}= \frac{a^2+b^2 m^2}{\left(a^2b^2-b^2 x_A{}^2-a^2y_A{}^2\right) \left(1+m^2\right)} [/itex]

so that

[itex]\frac{1}{\text{AP} \text{AP}'}+\frac{1}{\text{AQ} \text{AQ}'}=\frac{a^2+b^2}{a^2b^2-b^2 x_A^2-a^2y_A{}^2} [/itex] and this last expression doesn't actually depend on the coefficient

**". Link is here (no solution file available in the site for that year).***Annual Iowa Collegiate Mathematics Competition*The text is rather simple:

**Let E be an ellipse in the plane and let A be a fixed point inside of E. Suppose that two perpendicular lines through A intersect E in points P, P' and Q, Q' respectively. Prove that**

[itex]\frac{1}{\bar{AP} \cdot \bar{AP'}}+\frac{1}{\bar{AQ} \cdot \bar{AQ'}} [/itex]

is independent of the choice of lines.[itex]\frac{1}{\bar{AP} \cdot \bar{AP'}}+\frac{1}{\bar{AQ} \cdot \bar{AQ'}} [/itex]

is independent of the choice of lines.

__inscribed__into a generic ellipse and the property to demonstrate involve the four segments in which the diagonals are reciprocally subdivided.Till now I've found that:

- the property is actually true. I've seen that by empirical verification using
*Geogebra*and following the cartesian analytic solution (see^{1}) outlined below; - using a cartesian reference system it's possible to prove the property, but with
calculations (see*huge*^{1}); - drawing the tangent lines to the ellipse in the points P, P', Q, Q' they intersect in four points (say B, C, D, F). The diagonals of this new quadrilateral (this time
__circumscribed__to the ellipse) meets too in the point A (**!**) (empirical but unproven discover using*Geogebra*- probably, the core of the*yet unfound*geometrical demonstration is connected to this fact).

Since I'm not satisfied with the analytical cartesian way (with huge calculations) and I don't think that was the proof intended by the people that proposed this problem back in 1995 and since I think there must be a clever and a more elegant geometrical way to it, I'm asking for

**some help**to find a better solution.I've also challenged my math students (aged 16-18, about 70 students) to find a solution to this problem, and I'm curious to see if a fresher (and just less experienced) brain than mine can find the right way when my

*purported*experience isn't much helpful in this case.^{1}Given the ellipse centered in the origin with equation [itex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/itex], a generic point [itex]A\left(x_A,y_A\right)[/itex]__inside__the ellipse (that is with [itex]\frac{x_A{}^2}{a^2}+\frac{y_A{}^2}{b^2}<1[/itex]) and the two perpendicular lines*r*: [itex]y=m\left(x-x_A\right)+y_A[/itex] and*s*: [itex]y=-\frac{1}{m}\left(x-x_A\right)+y_A[/itex] incident in*A*that intersect the ellipse respectively in the points P, P and Q, Q', after calculating the coordinates of the intersection points P, P', Q and Q' of*r*and*s*with the ellipse and the lengths of the segments AP, AP', AQ, AQ', then it is[itex]\frac{1}{\bar{AP} \cdot \bar{AP'}}= \frac{b^2+a^2 m^2}{\left(a^2b^2-b^2 x_A{}^2-a^2y_A{}^2\right) \left(1+m^2\right)} [/itex]

and

[itex]\frac{1}{\text{AQ} \text{AQ}'}= \frac{a^2+b^2 m^2}{\left(a^2b^2-b^2 x_A{}^2-a^2y_A{}^2\right) \left(1+m^2\right)} [/itex]

so that

[itex]\frac{1}{\text{AP} \text{AP}'}+\frac{1}{\text{AQ} \text{AQ}'}=\frac{a^2+b^2}{a^2b^2-b^2 x_A^2-a^2y_A{}^2} [/itex] and this last expression doesn't actually depend on the coefficient

*, that is on the choice of the two lines***m***r*and*s*.
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