My Math Forum Difficult problem: deltoid inscribed into an ellipse

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 January 27th, 2013, 03:53 AM #1 Newbie   Joined: Jan 2013 Posts: 1 Thanks: 1 Difficult problem: deltoid inscribed into an ellipse I'm a math 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 "Annual Iowa Collegiate Mathematics Competition". Link is http://mathcs.central.edu/MAA/contes...ms/Probs95.htm (no solution file available in the site for that year). 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 $\frac{1}{\bar{AP} \cdot \bar{AP'}}+\frac{1}{\bar{AQ} \cdot \bar{AQ'}}$ is independent of the choice of lines. Practically there's a "deltoid" (a quadrilateral with perpendicular diagonals) 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;[/*:m:2nbm9j8n] using a cartesian reference system it's possible to prove the property, but with huge calculations (see (1) );[/*:m:2nbm9j8n] 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).[/*:m:2nbm9j8n] 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 $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, a generic point $A\left(x_A,y_A\right)$ inside the ellipse (that is with $\frac{x_A{}^2}{a^2}+\frac{y_A{}^2}{b^2}<1$) and the two perpendicular lines r: $y=m\left(x-x_A\right)+y_A$ and s: $y=-\frac{1}{m}\left(x-x_A\right)+y_A$ 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 $\frac{1}{\bar{AP} \cdot \bar{AP'}}= \frac{b^2+a^2 m^2}{\left(a^2b^2-b^2 x_A{}^2-a^2y_A{}^2\right) \left(1+m^2\right)}$ and $\frac{1}{\text{AQ} \text{AQ}'}= \frac{a^2+b^2 m^2}{\left(a^2b^2-b^2 x_A{}^2-a^2y_A{}^2\right) \left(1+m^2\right)}$ so that $\frac{1}{\text{AP} \text{AP}'}+\frac{1}{\text{AQ} \text{AQ}'}=\frac{a^2+b^2}{a^2b^2-b^2 x_A^2-a^2y_A{}^2}$ and this last expression doesn't actually depend on the coefficient m, that is on the choice of the two lines r and s. Thanks from Greek3
 August 7th, 2014, 08:00 AM #2 Newbie   Joined: Aug 2014 From: Milan Posts: 2 Thanks: 0 Could you write all the passages that led you to the solution??? Thanks a lot!!!
 August 10th, 2014, 10:35 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,629 Thanks: 2077 This is more readable.

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