# figure of eight- need help

• Nov 26th 2008, 11:35 AM
sonia1
figure of eight- need help
Hi. Does anyone know

'What is the largest figure of eight that will stand inside an equilateral triangle of side length one?' (the circles are of the same size).

I have no idea how to start it http://static.thestudentroom.co.uk/i...lies/frown.gif
• Dec 2nd 2008, 02:57 PM
Soroban
Hello, sonia1!

A fascinating problem . . .

Quote:

What is the largest figure of eight that will stand inside
an equilateral triangle of side length one?
(The circles are of the same size.)

Code:

                  A                   *                 /|\                 / | \               /  *  \               /*  |  *\             *  r|    *D             *    *O    *           / *  r|    * \           /    *  |  *    \         /        *        \         /      *  |  *      \       /    *  r|    *    \       /    *    *    *    \     /      *  r|    *      \     /          *  |  *          \  B *  - - - - - - * - - - - - - - * C

We have equilateral triangle $ABC$ of side 1.

Two equal circles are "stacked" inside.
Let their radius be $r.$
Let $O$ be the center of the upper circle.

Draw $OD \perp AC$.
Then $OD = r$

In right triangle ODA, we have: . $\angle OAD = 30^o\text{ and }OD = r$
. . Then: . $OA = 2r$

Hence, the height of the triangle is $5r.$

But the height of an equilateral triangle of side 1 is: . $\frac{\sqrt{3}}{2}$

Therefore: . $5r \:=\:\frac{\sqrt{3}}{2} \quad\Rightarrow\quad\boxed{ r \:=\:\frac{\sqrt{3}}{10}}$

• Dec 3rd 2008, 09:51 AM
sonia1
thank you
Hi thank you so much for your help, i'm not good at doing geometry problems.

I'm still stuck on two other geometry problems:

1. is it possible to have an equilateral triangle in the plane all of whose vertices lie on the integer lattice{(m,n)|m,n are integers}? If so, what is the shortest side lengh possible for such a triangle? what about other regular polygons?

2. Is it possible for the tangents to a parabola at two distinct zeroes (x intercepts) to meet at right angles? Is it possible to find such a parabola for which the x and y intercepts and the point of intersection of the two tangents lie on the integer lattice? What can you say about cubics?

do you think you can help?
• Dec 3rd 2008, 06:21 PM
Soroban
Hello, sonia1!

I have part of #2 . . .

Quote:

2. Is it possible for the tangents to a parabola at two distinct zeroes (x intercepts)
to meet at right angles?

Any parabola can be translated so its y-intercept is on the y-axis.

Code:

              |   *          |          *   \          |          /     *        |        * - - -*- - - - + - - - -*- - -     -c\*      |      */c       \  *  |  *  /         \    *    /         \    |    /           \  |  /           \  |  /             \ | /             \|/               *               |

The equation of this parabola is: . $y \:=\:a(x^2 - c^2)$

The intercepts are: $(c, 0),\;(\text{-}c, 0)$

The slope of the tangent is: . $y' \:=\:2ax$

. . At $x = c$, the slope is: . $m_1 \:=\:2ac$

. . At $x = \text{-}c$, the slope is: . $m_2 \:=\:\text{-}2ac$

If these slopes are perpendicular: . $m_1 \:=\:-\frac{1}{m_2}$

So we have: . $2ac \:=\:-\frac{1}{\text{-}2ac} \quad\Rightarrow\quad 4a^2c^2\:=\:1 \quad\Rightarrow\quad c^2\:=\:\frac{1}{4a^2}$

. . Hence: . $c \:=\:\pm\frac{1}{2a}$

The equation of the translated parabola is: . $y \;=\;a\left(x^2 - \frac{1}{4a^2}\right)$

• Dec 6th 2008, 02:33 PM
sonia1
hey thanks for your help I really appreciate it.

I just wanted to ask, you know for the figure of eight problem what if the circles are next to each other horizontally and not standing up inside
an equilateral triangle of side length one?, then how would you work that out?

• Dec 9th 2008, 08:50 AM
Laurent
Quote:

Originally Posted by sonia1
2. Is it possible for the tangents to a parabola at two distinct zeroes (x intercepts) to meet at right angles? Is it possible to find such a parabola for which the x and y intercepts and the point of intersection of the two tangents lie on the integer lattice? What can you say about cubics?

do you think you can help?

With Soroban's notations, the x-intercepts are $\left(\pm\frac{1}{2a},0\right)$, the y-intercept is $\left(0,-\frac{1}{4a}\right)$ (substitute $x=0$ in the equation of the parabola), and the intersection of the tangents is $\left(0,-\frac{1}{2a}\right)$. As soon as $\frac{1}{4a}$ is an integer (for instance, if $a=\frac{1}{4}$), all these points lie on the integer lattice.
• Dec 9th 2008, 10:47 AM
Opalg
Quote:

Originally Posted by sonia1
I just wanted to ask, you know for the figure of eight problem what if the circles are next to each other horizontally and not standing up inside an equilateral triangle of side length one?, then how would you work that out?

$\setlength{\unitlength}{2mm}
\begin{picture}(20,20)
\put(6.5,4.6){\circle{15}}
\put(13.5,4.6){\circle{15}}
\put(0.1,1){\line(1,0){19.8}}
\put(10,17.4){\line(-3,-5){9.8}}
\put(10,17.4){\line(3,-5){9.8}}
\put(10,1){\line(0,1){16.4}}
\put(10,1){\line(1,1){3.5}}
\put(13.5,1){\line(0,1){3.5}}
\qbezier(13.5,4.5)(13.5,4.5)(19.8,1)
\end{picture}$

From the picture, you should be able to see that if the radius of the circles is r, then $r(1+\sqrt3) = 1/2$ and so $r = \frac{\sqrt3-1}4\approx0.183$. This is greater than $\sqrt3/10\approx0.173$, so that positioning beats Soroban's! (Clapping)

Intuitively, it looks as though having the two circles alongside each other like that must be the optimum configuration, but I wouldn't want to have to prove that.