Polygons And Triangles

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Polygons And Triangles

page revision: 0, last edited: 15 Sep 2008 08:54

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Help on Ex. 55

Hello there!

I am from the Czech Republic and thus my Math education was nearly completely proof-less. Thus I have started reading How to Prove It by Velleman for general proofs and Kiselev's Geometry to brush up my planimetry / stereometry (which I lack intention for and generally I'm pretty bad at, even tough the rest of high school math is easy for me and I'm want to study Math at a university) and to learn how to do geometrical proofs.

I'm stuck on exercise 55. I have tried to work with the definitions from the preceding chapter but to no avail. It maybe because I have nearly zero experience with geometrical proofs, so any tips and tricks would be highly appreciated.

So far I have tried to proof this by cases (ie. convex / non convex, complex / non complex), but I don't know how to continue.

I feel really ashamed as this is Grade 7 material that I cannot do in the last year of high school.

ReplyOptionsProve that each diagonal of a quadrilateral either lies entirely in its interior, or entirely in its exterior. Give an example of a pentagon for which this is false.First, you should stop feeling ashamed: with geometry, beginners experience the same difficulties, regardless of age or grade.

Second, you should stop using the expression "do proofs". There is no such an activity in mathematics as "doing proofs": we are concerned not with how to derive some things from other things, but what is true in the world around us, and what is false, and why.

In this particular problem, I would start with constructing a counter-example for pentagons.

Then I would try to construct a counter-example for quadrilaterals (even though the problem claims they don't exist), for - who knows maybe the book is wrong? After some tries, if you don't succeed, maybe you'll get some idea why. Presenting such an idea in a form neat enough to guarantee that there are no

loopholes, would constitute a solution to the problem.

Regards,

Alexander

ReplyOptionsThanks for creating and maintaining this site. I expect to use it often!

I would be grateful if you could repond to my answer to question 55

Prove that each diagonal of a quadrilateral either lies entirely in its interior, or entirely in its exterior. Give an example of a pentagon for which this is false.

1. for any vertex in any polygon, if the angle between a diagonal and a side is greater than the angle between the two sides

then the diagonal lies in the exterior of the area bounded by the two sides

2. for any vertex in any polygon, if the angle between a diagonal and a side is less than the angle between the two sides

then the diagonal lies within the interior of the area bounded by the two sides

3. for any vertex in a convex quadrilateral, the angle between a diagonal and a side is less that the angle between the sides

4. for two vertices in a quadrilateral that is not convex, the angle between a diagonal and a side is greater than the angle between the sides

5. a quadrilateral is either convex or not convex

6. therefore, the diagonal in a quadrilateral either lies entirely within the interior of the quadrilateral or entirely within the exterior of the quadrilateral.

ReplyOptionsDear Matthew,

I'd suggest that you start with constructing the required counter-example of a pentagon, which has a diagonal lying entirely neither entirely inside or entirely outside the pentagon.

Then you will see that your argument about quadrilaterals cannot be correct, since at no point were you using the fact that the polygon is a quadrilateral, not a pentagon.

Also, note that your sequence of statements is not really an "argument": you make claims, but don't even try to explain why they are correct

[In fact some of them are incorrect as stated, e.g. 2].

ReplyOptionsExercise 32:

Four points on the plane are vertices of three different quadrilaterals. How can this happen?Construct the four points in the plane by placing 3 in a triangular formation with the fourth in the center of the others. 3 quadrilaterals can be constructed by alternating from which two of the outer 3 points the sides of the fourth(center) point connect.

ReplyOptionsExercise 53:

Can a convex broken line self-intersect?No. The self-intersection would put part of the broken line on both sides of another of its segments. This violates the definition in $\S 31$.

$\S 31$) A convex broken line lies on one side of each of ist segments continued indefinitely in both directions.

ReplyOptionsExercise 54:

No:Is it possible to tile the entire plane by non-overlapping polygons all of whose angles contain $140^{\circ}$ each?ReplyOptionsExercise 55:

Prove that each diagonal of a quadrilateral either lies entirely in its interior, or entirely in its exterior. Give an example of a pentagon for which this is false.Given quadrilateral $ABCD$ choose any vertex, e.g. $A$. This vertex belongs to the same sides as two other vertices, $B$ and $C$. Therefore the only diagonal is to the remaining vertex $D$, $\overline{AD}$.

The quadrilateral perimeter is: $\overline{AB}$, $\overline{AC}$, $\overline{CD}$, $\overline{DB}$. Vertex $A$ has sides $\overline{AB}$ and $\overline{AC}$ emanating from it. The diagonal $\overline{AD}$ cannot intersect either of these two sides as that would mean $\overline{AD}$ does not exist(by $\S 4$ it would share 2 points[A and the point of intersection] and be along the same line, further this invalidates D being a vertex and there being a quadrilateral).

If diagonal $\overline{AD}$ intersected the remaining quadrilateral sides($\overline{BD}$ or $\overline{CD}$) other than at point $D$, $\overline{AD}$ and the side would share this intersection point and point $D$. By $\S 4$ the diagonal and the side would then be along the same line, invalidating $D$ as a vertex and the existence of the quadrilateral.

A diagonal emanating from vertex $A$ originates interior or exterior to the quadrilateral. For it to lie both exterior and interior it would need to cross through(intersect) a side of the quadrilateral. But it has been shown that diagonal $\overline{AD}$ cannot intersect any side of the quadrilateral. Thus a diagonal of a quadrilateral lies exclusively interior or exterior to the quadrilateral.

Example of a diagonal in a pentagon which lies both interior and exterior to the pentagon:ReplyOptionsExercise 56:

Prove that a closed convex broken line is the boundary of a polygon.$\S 32$) Polygon is a figure formed by a non-self-intersecting closed broken line together with the part of the plane bounded by the line. Boundary is defined as the broken line itself.

$\S 31$) A broken line is convex if it lies on one side of each of its segments continued indefinitely in both directions.

A convex broken line cannot self-intersect as this would violate the definition of being convex, the broken line lying entirely on one side of any of its continued segments. Therefore the closed convex broken line satisfies the definition of a polygon boundary.

ReplyOptionsExercise 57:

Is an equilateral triangle considered isoceles? Is an isoceles triangle considered scalene?$\S 33$ defines an isoceles triangle as having 2 congruent sides. It does not strictly state "only" 2 congruent sides. An equilateral being defined as having 3 congruent sides can be considered isoceles.

$\S 33$ defines a scalene triangle as having 3 sides of differing length. Therefore an isoceles triangle, which has 2 sides of the same length, cannot be scalene.

ReplyOptionsExercise 58:

How many intersection points can three straight lines have?$\S 4$ Any two straight lines can intersect at most at 1 point. Let the lines be $A$, $B$, $C$.

All intersection combinations are: $AB$, $AC$, $BA$, $BC$, $CA$, $CB$.

Removing redundancies: $AB$, $AC$, $BC$.

Three straight lines can intersect at as many as 3 points.

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