Cool Problem 2.0 (1 Viewer)

Carrotsticks · May 9, 2012

Maybe I can have a 'Cool Problem of the Day' sorta thing going...

A right angled triangle is inscribed within a circle such that all 3 vertices are on the circumference of the circle.

Given the ratio...

$\frac{A_{circle}}{A_{triangle}} = 2 \pi$

...where A_x denotes the area of x, find the other two angles of the triangle.

Carrotsticks · May 9, 2012

RealiseNothing · May 9, 2012

I got ~18 and 72.

Carrotsticks · May 9, 2012

RealiseNothing said:
I got ~18 and 72.

Close! But no cigar...

The solutions are integers.

RealiseNothing · May 9, 2012

Carrotsticks said:
Close! But no cigar...

The solutions are integers.

I ended up with one of my angles being $tan^{-1}(\frac{1}{3})$

The other angle was just it's complement.

I'll have another go if that's not right later (friend's over).

funnytomato · May 9, 2012

Carrotsticks said:
Close! But no cigar...

The solutions are integers.

15°， 75°
is this^ right
???

Carrotsticks · May 9, 2012

Yep! Would you like to explain how you got your answer? (or maybe wait until a few more people try).

RealiseNothing · May 9, 2012

Just retried and got the same as funnytomato, will post up my solution.

I realise what I did wrong now.

RealiseNothing · May 9, 2012

Here's my proof, if you don't want to see it, skip this post.

The area of a circle is $\pi r^2$ , and thus $\frac{\pi r^2}{2\pi} = ~area ~of~ triangle = \frac{r^2}{2}$

We know that the length of the hypotenus is $2r$ , and thus the perpendicular height of the triangle is $\frac{r}{2}$ .

So we know that the perpendicular distance of the right angle to the hypotenuse(diameter) is $\frac{r}{2}$ , hence if we construct a triangle inside the original triangle using this line as a side, we can find the angles of the original triangle. As shown below, the blue line is the line we added to construct a smaller triangle.

If we find the length of the other side of this smaller triangle, we can use trigonometry to find the angles of the original triangle. Below is a "cut-out" of the circle, and magnified to show the side we are trying to find, along with the already known sides.

Hence the base of that triangle is $\sqrt{r^2 - \frac{r^2}{4}} = \frac{r\sqrt3}{2}$

So we can now put in this value to our original diagram:

To find theta, we just use trig:

$tan\theta = \frac{\frac{r}{2}}{r+\frac{r\sqrt3}{2}}$

After some simplification:

$\theta = tan^{-1}(\frac{1}{2+\sqrt3})$

$\theta = 15$

Hence the angles are 15 and 75.

FINISHED, YOU CAN LOOK NOW

barbernator · May 9, 2012

woah, some people overcomplicated the solution. Again, if you haven't solved it, dont look.

$\frac{\pi r^2}{0.5ab}=2\pi \\ \frac{r^2}{ab}=1\\ r^2=ab\\ but~a=2rsin(\theta ),~and~b=2rcos(\theta)\\ \therefore r^2=4r^2sin(\theta )cos(\theta )\\ 1=2sin(2\theta )\\ sin(2\theta )=\frac{1}{2}\\ 2\theta =\pi /6\\ \theta =15~degrees,~\alpha =75~degrees.$

a qualifier As the triangle is right angled, its hypotenuse is the diameter of the circle.

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Cool Problem 2.0 (1 Viewer)

Carrotsticks

Retired

Carrotsticks

Retired

RealiseNothing

what is that?It is Cowpea

Carrotsticks

Retired

RealiseNothing

what is that?It is Cowpea

funnytomato

Active Member

Carrotsticks

Retired

RealiseNothing

what is that?It is Cowpea

RealiseNothing

what is that?It is Cowpea

barbernator

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