Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**sydbernard****Member**- Registered: 2017-09-04
- Posts: 6

Construct a triangle ABC given its side AB, the sum of the other two sides AC + CB and its angle at C.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

hi sydbernard

It is a property of all ellipses that the distance from any point on the ellipse to one focus added to the distance to the other, is a constant. It is a property of all circles that if A, B and F are all points on the circumference then angle AFB = constant for all F on the same arc from A to B. My construction uses these two properties.

First some theory for ellipses.

The Cartesian equation for an ellipse is

Where 2a is the length of the major axis and 2b the length of the minor axis.

For this problem I'm labelling the foci (plural of focus) A and B. So AB will be the given length and AC + CB will be a specified constant amount.

I'll illustrate with AB = 8, AC + CB = 10 and ACB = 30. The method will work for any such measurements but these make the construction easier to follow.

On a coordinate diagram mark the centre of the ellipse as O, and put A and B equal distances (4 units) either side of O.

If point Y is chosen to be on the ellipse and directly above O then, as AO = 4 and AY = 5, this means that OY = 3 by Pythagoras.

So we know the minor axis has b = 3.

At the moment we don't know the major axis but it is easy to calculate using the constant distance property. Let the major axis be DE.

Then AD + DB = 10, so a = 5. So the equation of the ellipse is

So C has to be somewhere on that ellipse.

Now mark a point F above the line AB, so that angle AFB = 30. I want a circle to go through A, B and F. Clearly the centre has to be on the Y axis. I've bisected AF (point G) and drawn a line through G perpendicular to AF, to cross the Y axis at H. This is the centre of the required circle. I drew the circle and where it crosses the ellipse is the point C.

I used Geometers Sketchpad to make the construction but you can also use Geogebra.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

Further thoughts:

Constructing a 30 degree angle is fairly straight forward but what if it's another angle, θ. Find F by (1) Drawing a line straight up from A; (2) Draw a line from B,

angle 90 - θ; (3) where these lines cross is F with AFB = θ

Some other ways of drawing the ellipse. We're used to using a compass to make a circle or an arc. There's a similar way to make an ellipse. Tie a piece of string in a loop exactly AB + BC + CA long. Put two drawing pins into the paper at the focal points. Loop the string around both pins and pull tight with a pencil. Allow the pencil to track around the foci and it will draw an ellipse.

There is also a construction which fixes points on the perimeter of the ellipse. Then you join them with a smooth curve. It's about 50 years since I last used this technique so I'll have to do some hard thinking if you want to see how it's done

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**