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 Post subject: Why can't I get this...?
PostPosted: Fri, 4 May 2012 23:19:01 UTC 
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Given a rectangle X_1X_2X_3X_4, construct circles O_1,O_2,O_3,O_4 such with centers X_1,X_2,X_3,X_4 with radii r_1, r_2, r_3, r_4respectively, such that: r1+r3=r2+r4<d, (d is the length of the diagonal of rectangle X_1X_2X_3X_4). Construct the common external tangents of circles O_1 and O_3. Construct the common external tangents of circles O_2 and O_4. These 4 lines form a quadrilateral. Show that one can inscribe a circle into this quadrilateral.

My Work:
Well the only thing that I know about circles being inscribed in quadrilaterals, is that we can inscribe a circle in a quadrilateral ABCD iff:
AB+CD=BC+DA,

I have no idea how to use the fact that r_1+r_3=r_2+r_4

At first I thought, "hmmmm the r_1+r_3=r_2+r_4 condition looks suspiciously like AB+CD=BC+DA,"

So, I thought similar triangles should destroy this problem. But, I couldn't find any similar triangles that could relate to the sides of the quadrilateral that we construct.

I feel like I'm being really dumb... >_<

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PostPosted: Sat, 5 May 2012 08:16:05 UTC 
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rdj5933mile5math64 wrote:
Given a rectangle X_1X_2X_3X_4, construct circles O_1,O_2,O_3,O_4 such with centers X_1,X_2,X_3,X_4 with radii r_1, r_2, r_3, r_4respectively, such that: r1+r3=r2+r4<d, (d is the length of the diagonal of rectangle X_1X_2X_3X_4). Construct the common external tangents of circles O_1 and O_3. Construct the common external tangents of circles O_2 and O_4. These 4 lines form a quadrilateral. Show that one can inscribe a circle into this quadrilateral.

My Work:
Well the only thing that I know about circles being inscribed in quadrilaterals, is that we can inscribe a circle in a quadrilateral ABCD iff:
AB+CD=BC+DA,

I have no idea how to use the fact that r_1+r_3=r_2+r_4

At first I thought, "hmmmm the r_1+r_3=r_2+r_4 condition looks suspiciously like AB+CD=BC+DA,"

So, I thought similar triangles should destroy this problem. But, I couldn't find any similar triangles that could relate to the sides of the quadrilateral that we construct.

I feel like I'm being really dumb... >_<


Maybe you can try to prove directly that the centre of the rectangle is equidistance from all four sides of the quadrilateral, which follows trivially from ...

Then prove that indeed the feet of the perpendiculars lie inside those four sides.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sat, 5 May 2012 17:34:28 UTC 
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outermeasure wrote:
Maybe you can try to prove directly that the centre of the rectangle is equidistance from all four sides of the quadrilateral, which follows trivially from ...


r_1+r_3=r_2+r_4

outermeasure wrote:
Then prove that indeed the feet of the perpendiculars lie inside those four sides.

Darn which is also trivial. It's a little tricky to formalize but very intuitive.

Thank You outermeasure! :D

EDIT (a question): I take it that it was a good diagram that provided the intuition for your solution. Do you know where I might be able to find a good compass? Also, how do I get quicker at drawing good diagrams? I know basic Euclidean construction stuff, but it still takes me a long time to construct diagrams.

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PostPosted: Sat, 5 May 2012 18:45:32 UTC 
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Pictures are like any other skill, earned by hours of practice and repetition.

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PostPosted: Sun, 6 May 2012 00:36:44 UTC 
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Shadow wrote:
Pictures are like any other skill, earned by hours of practice and repetition.


Thank you for the words of wisdom! :D

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PostPosted: Sun, 6 May 2012 18:30:34 UTC 
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rdj5933mile5math64 wrote:
I take it that it was a good diagram that provided the intuition for your solution. Do you know where I might be able to find a good compass? Also, how do I get quicker at drawing good diagrams? I know basic Euclidean construction stuff, but it still takes me a long time to construct diagrams.


I don't think IMO allows compasses. Of course, you can use your glass to draw a circle in IMO :P

Actually I was looking at ways to use the condition r_1+r_3=r_2+r_4 which points me to the diagonals and the centre of the rectangle, as they are the only apparent place to look where the equality manifests itself.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sun, 6 May 2012 20:34:16 UTC 
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outermeasure wrote:
I don't think IMO allows compasses. Of course, you can use your glass to draw a circle in IMO :P


Lol IMO allowed them last year. If you don't mind me asking, have you done the IMO before?

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PostPosted: Mon, 7 May 2012 10:28:42 UTC 
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rdj5933mile5math64 wrote:
outermeasure wrote:
I don't think IMO allows compasses. Of course, you can use your glass to draw a circle in IMO :P


Lol IMO allowed them last year. If you don't mind me asking, have you done the IMO before?


No, I am not a former IMO contestant.

And I'm pretty sure that the IMO didn't allow compasses a decade ago (I heard the idea of using the glass of water to draw circles from a friend of mine who did IMO back then), although national olympiads always do. They must have changed the regulation a few years ago.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 7 May 2012 13:49:36 UTC 
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outermeasure wrote:
No, I am not a former IMO contestant.


Neither am I - and I don't think that I ever will be unless my parents decide to move to Guam or Puerto Rico lol (I have terrible luck with AIMEs - last year (first time I heard of the AIME exam) one of the questions was misworded which cost me an hour and this year I made 4 dumb mistakes.).

outermeasure wrote:
Actually I was looking at ways to use the condition r_1+r_3=r_2+r_4 which points me to the diagonals and the centre of the rectangle, as they are the only apparent place to look where the equality manifests itself.


Thanks for the intuition! :D

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