Hard geometry questions

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

Hard geometry questions are posted here daily:

http://www.8foxes.com/

They may be useful for those teaching geometry.

**Posted:** Thu Mar 12, 2009 10:10 am

Altheman

hmm...your posts always look like such spam but there are actually some really nice problems on your site (!)

**Posted:** Thu Mar 12, 2009 11:13 am

_________________
-Alex
Altheman's Problem Column

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

Hello Altheman,
The questions at the site are all original.
And open to everyone. You don't need to get a permission.

By the way, I've checked your blog. That's impressive too.

polarfox

I like #39 very much:
http://www.8foxes.com/

**Posted:** Sun Mar 15, 2009 8:03 pm

Brut3Forc3

Is that quadrilateral a square? (Or a rectangle or neither?)

**Posted:** Sun Mar 15, 2009 9:53 pm

_________________
"I can stand brute force, but brute reason is quite unbearable. There is something unfair about its use. It is hitting below the intellect." - Oscar Wilde

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

It was intended for a square at first.
But the answer is the same for a rectangle as well.

AND, I am not sure but I think the answer should not change if it is a just a parallelogram, which makes it beautiful. That's what I call geometric esthetics.

polarfox

**Posted:** Mon Mar 16, 2009 6:35 am

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

Although there is a nice result for parallelogram, the answer is different. I stand corrected.
But the questions is valid for any rectangle and any circle. Their sizes don't matter. There relative positions don't matter. The circle is in or out of the rectangle don't matter. As long as all 4 corners of the rectangle is out of the circle. The answer is the same.

-Polarfox

**Posted:** Fri Mar 20, 2009 9:31 pm

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

$x^2 + 2 = 12 + 14$
should give the answer.

**Posted:** Wed Apr 01, 2009 6:31 pm

mathwizarddude

Does the website contain answers/solutions to the problems?

**Posted:** Wed Apr 01, 2009 7:29 pm

Sephiroth

hmm it looks like british flag theorem, but the tangents are coming from different points. yes the result is the same. hmm...

**Posted:** Wed Apr 01, 2009 7:34 pm

_________________
"Ask and it will be given to you; seek and you will find; knock and the door will be opened to you. For everyone who asks receives; he who seeks finds; and to him who knocks, the door will be opened." -Matthew 7:7-8

grn_trtle

How do you use British Flag theorem when they don't meet at a point but at two points on a circle?

**Posted:** Wed Apr 01, 2009 7:37 pm

_________________
$\arctan\alpha + \arctan\beta = \text{arg}\{(1+\alpha i)(1+\beta i)\} = \arctan\left( \frac{\alpha+\beta}{1-\alpha\beta} \righ...$

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

Never heard British Flag Theorem before. I knew the relationship but never knew that it had a name.
But you don't have to use it to find the solution.
Website has a stepwise-approach to hard problems. Solving #38 makes solving #39 easier.
Look at the picture below:

#39 is just one Pythagoras away from the solution to the above one! (OR should I say 4 Pythagoras away Smile

Can you see?

-polarfox

**Posted:** Sun Apr 05, 2009 9:36 pm

grn_trtle

Ok, yeah, 38 is pure British Flag theorem, plain and simple.

$15+12=2+x^2\implies x^2=25\implies x=\boxed{\textbf{(C)}\ 5}$ .

**Posted:** Sun Apr 05, 2009 9:55 pm

_________________
$\arctan\alpha + \arctan\beta = \text{arg}\{(1+\alpha i)(1+\beta i)\} = \arctan\left( \frac{\alpha+\beta}{1-\alpha\beta} \righ...$

mathwizarddude

For #39, doesn't the line segment formed by the two points of intersection on the circle have to be parallel to the verticle side of the rectangle?

**Posted:** Tue Nov 03, 2009 4:57 pm

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

No, it doesn't have to be parallel to any side of the square. AND it is not important! What is needed is that radiuses connected to the tangent points are perpendicular to the tangent lines. Then you can write Pithagoras for 4 tangent lines, which should give:
$15+12=2+x^{2}\implies x^{2}=25\implies x=\boxed{\textbf{(C)}\ 5}$

Also check the similar ones:
http://www.8foxes.com/Home/162

http://www.8foxes.com/Home/163

**Posted:** Wed Nov 04, 2009 9:46 pm

mathwizarddude

PQ parallel to AD

**Posted:** Wed Nov 04, 2009 10:23 pm

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

Hello mathwizard, here is your mistake:
"and let the two line segments from $A$ and $B$ intersect the circle at $P$ and the two line segments from $C$ and $D$ intersect the circle at $Q$ "

Line segments from $A$ and $B$ intersect the circle at TWO DISTINCT POINTS, not one. Similarly,
Line segments from $C$ and $D$ intersect the circle at TWO DISTINCT POINTS, not one.

That is leading you an inaccurate result.
A real-monster has just been posted. It is ugly:
http://www.8foxes.com/

**Posted:** Thu Nov 05, 2009 11:58 pm

mathwizarddude

Whoops, I didn't realize the four lines are supposed to be tangent to the circle (and thus must be four distinct points)... By the way do you have a solution page to those geometry problem? Your website is really awesome, by the way! Keep up the good work. (Also is there another one on other areas other than geometry?)

**Posted:** Fri Nov 06, 2009 4:41 pm

Eulers_Apprentice · Hodge Conjecture Offline Joined: 05 Aug 2008 Posts: 51

Polarfox, are you the owner of the website?

**Posted:** Fri Nov 06, 2009 8:39 pm

polarfox · New Member Offline Joined: 12 Mar 2009 Posts: 10

mathwizarddude,
Some answers are discussed and posted on: http://8foxes.blogspot.com/

Eulers_Apprentice, yes I own the website. It is a simple, stupid hobby.
You don't have a problem whit that, right? Mr. Green

-polarfox

**Posted:** Fri Nov 06, 2009 10:03 pm