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
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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
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 $\displaystyle ABC$ of side 1.
Two equal circles are "stacked" inside.
Let their radius be $\displaystyle r.$
Let $\displaystyle O$ be the center of the upper circle.
Draw $\displaystyle OD \perp AC$.
Then $\displaystyle OD = r$
In right triangle ODA, we have: .$\displaystyle \angle OAD = 30^o\text{ and }OD = r$
. . Then: .$\displaystyle OA = 2r$
Hence, the height of the triangle is $\displaystyle 5r.$
But the height of an equilateral triangle of side 1 is: .$\displaystyle \frac{\sqrt{3}}{2}$
Therefore: .$\displaystyle 5r \:=\:\frac{\sqrt{3}}{2} \quad\Rightarrow\quad\boxed{ r \:=\:\frac{\sqrt{3}}{10}}$
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?
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 yintercept is on the yaxis.
Code:
*  *
\  /
*  *
  *    +    *  
c\*  */c
\ *  * /
\ * /
\  /
\  /
\  /
\  /
\/
*

The equation of this parabola is: .$\displaystyle y \:=\:a(x^2  c^2)$
The intercepts are: $\displaystyle (c, 0),\;(\text{}c, 0)$
The slope of the tangent is: .$\displaystyle y' \:=\:2ax$
. . At $\displaystyle x = c$, the slope is: .$\displaystyle m_1 \:=\:2ac$
. . At $\displaystyle x = \text{}c$, the slope is: .$\displaystyle m_2 \:=\:\text{}2ac$
If these slopes are perpendicular: .$\displaystyle m_1 \:=\:\frac{1}{m_2}$
So we have: .$\displaystyle 2ac \:=\:\frac{1}{\text{}2ac} \quad\Rightarrow\quad 4a^2c^2\:=\:1 \quad\Rightarrow\quad c^2\:=\:\frac{1}{4a^2}$
. . Hence: .$\displaystyle c \:=\:\pm\frac{1}{2a}$
The equation of the translated parabola is: .$\displaystyle y \;=\;a\left(x^2  \frac{1}{4a^2}\right)$
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?
With Soroban's notations, the xintercepts are $\displaystyle \left(\pm\frac{1}{2a},0\right)$, the yintercept is $\displaystyle \left(0,\frac{1}{4a}\right)$ (substitute $\displaystyle x=0$ in the equation of the parabola), and the intersection of the tangents is $\displaystyle \left(0,\frac{1}{2a}\right)$. As soon as $\displaystyle \frac{1}{4a}$ is an integer (for instance, if $\displaystyle a=\frac{1}{4}$), all these points lie on the integer lattice.
$\displaystyle \setlength{\unitlength}{2mm}From the picture, you should be able to see that if the radius of the circles is r, then $\displaystyle r(1+\sqrt3) = 1/2$ and so $\displaystyle r = \frac{\sqrt31}4\approx0.183$. This is greater than $\displaystyle \sqrt3/10\approx0.173$, so that positioning beats Soroban's! (Clapping)
\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}$
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.