Construction problems

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Geometry original article by student Dan Leonte

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Mathematical articles and notes   Abstract: Starting from a geometry theorem, this material presents an original method / strategy for solving construction problems. Keywords: geometry, construction problem Date: the 7th of June, 2013 Author: Dan Leonte, student, National College << Mihai Eminescu>> Botosani, Romania:

Mathematical articles and notes Abstract: Starting from a geometry theorem, this material presents an original method / strategy for solving construction problems. Keywords : geometry, construction problem Date: the 7th of June, 2013 Author: Dan Leonte , student, National College << Mihai Eminescu >> Botosani , Romania

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The constructions considered here are done with an unmarked straightedge, only one side. That is we are allowed to : • Draw a line through two known points • Get knew known points from intersections of such drawn lines or with a previously drawn (=given) curve ( eg . circle). • Choose any arbitrary point, as long as the final construction doesn't depend on the choice. Full stop. Any other action is forbidden .

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Theorem Given two points A,B on a line l, and the midlepoint M of [AB] , a parallel to l by any exterior point is constructable using only the ruler/straightedge. Proof Be N an arbitrary exterior point from line l. I will construct the parallel to l passing through N. Using the straightedge I join points A and N. Be P a point on the half-line (AN over N. I join now points P and B, P and M, and , of course, N and B.

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Let {O}=NB ∩ PM. Now I draw AO. Be point {D}= AO ∩ BP. From Ceva ‘s Theorem we get AM/MB *DB/DP * PN/NA =1 But AM=MB => DB/DP=NA/NP   DN║AB , q.e.d .

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Using this theorem, a lot of problems can be proved. Here are some examples: 1. Having given a circle of center O, C(O) and a line (l) exterior to the circle, show that if two points are taken arbitrary on l, the middlepoint of segment [MN] is constructable using only a straightedge . 2.Be drawn a circle of center O,C(O), and a point {A} on circle C(O). By A it is drawn the tangent at C(O), and on that is taken point {M}.Show that the reflection of point M over A is constructable using only a straightedge. 3.Having drawn a parallelogram and a line l passing throw one of it’s vertices, only with the straightedge, show that it is constructable the middlepoint of the segment determined by the intersection of the line l with the sides of the parallelograms which does not pass through that vertices. 4.Having drawn a circle of center A, C(A), using only the straightedge, show that the tangent in a point {B} on the circle C(A) is constructable . Author of these problems : Nita Cristi

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In this article I will show a proof for the last problem, all the other problems having similar solutions to that. Of course, the theorem above is making our work much easier . Proof Using the rule, I draw the line AB. Be point {C} the intersection of this line with the circle C(A). Be D an arbitrary point on the circle. Using the lemma above, I can construct a trapezoid BCDG ( where {E}=CD ∩BG )and, the most important, the parallel to line BC passing through point D. Now, the picture looks like that:

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Using the straightedge, I extend segment [DG] over G. Be {H}=DG∩C(A). Joining points B and H and intersecting line BH with line CD we obtain point {I}. Now, I draw segment [IA]. Because of the fact that BHDC is a trapezoid inscribed in a circle, we get that BHDC is an issoscel trapezoid (1). It is well known that in a trapezoid, the middle -points of the parallel sides are collinear with the point of intersection of the diagonals, and, further more, with the point of intersection of the unparallel sides (2). From (2) we obtain that IA is the same line as the line which passes trough the middle-points of segments [BC] and [HD], but from (1) we got that BCDH is an issoscel trapezoid  IA ┴ BC, and, of course, IA ┴ HD.(3) Be {J} and {K} the points where line IA intersects the circle C(A ).

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Now, we have drawn a segment [JK], and, also, his mid-point, A  ( applying the lemma one more time) we can draw a parallel line passing throw B. Let this line be t. So, we have IA ┴ BC, IA║ t  t ┴ BC, but [BC] is diameter in circle C(A)   t is the tangent in point B, q.e.d .

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