Inscribed-Quad in an Excribed-Quad

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sowmitra
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Inscribed-Quad in an Excribed-Quad

Unread post by sowmitra » Sun May 03, 2015 2:35 pm

The quadrilateral $ABCD$ is excribed around a circle with centre $I$. Prove that, the projections of $B$ and $D$ on $IA$, $IC$ lie on a circle.

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Last edited by sowmitra on Mon May 04, 2015 8:38 pm, edited 1 time in total.
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Nirjhor
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Re: Inscribed-Quad in an Excribed-Quad

Unread post by Nirjhor » Mon May 04, 2015 1:28 am

You sure about the statement?
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


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sowmitra
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Re: Inscribed-Quad in an Excribed-Quad

Unread post by sowmitra » Mon May 04, 2015 8:39 pm

Extremely sorry for the typo... :oops:
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

Nirjhor
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Re: Inscribed-Quad in an Excribed-Quad

Unread post by Nirjhor » Mon May 04, 2015 9:43 pm

Image

Points $W,X$ are projections of $B$ and $Y,Z$ are projections of $D$ on $IA$ and $IC$ respectively. Multiple configs can occur, so angles are directed mod $\pi$.

Since $\measuredangle IWB=\measuredangle IXB=\pi/2$ quad $IWBX$ is cyclic. Similarly quad $IZDY$ is cyclic.

So we have \[\measuredangle WXI = \measuredangle WBI=\pi/2 - \measuredangle BIW=\pi/2+\measuredangle ABI+\measuredangle IAB=\frac{1}{2}\left(\pi+\measuredangle ABC+\measuredangle DAB\right)\] and similarly $\measuredangle IYZ=\frac{1}{2}\left(\pi+\measuredangle DCB+\measuredangle ADC\right)$.

Hence $WXYZ$ cyclic iff \[\measuredangle WXI=\measuredangle IYZ\Leftrightarrow \measuredangle ABC+\measuredangle DAB=\measuredangle DCB+\measuredangle ADC\] \[\Leftrightarrow \measuredangle ADC+\measuredangle DCB+\measuredangle CBA+\measuredangle BAD=0\] which is true for quad $ABCD$.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Tahmid
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Re: Inscribed-Quad in an Excribed-Quad

Unread post by Tahmid » Wed May 06, 2015 12:34 pm

i could not visit the forum last 2 days for internet troubles . so didn't notice the typo update .
was trying to solve the previous .....then hssss :cry:

whatever, now i got the solution which nirjhor posted :|

tanmoy
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Re: Inscribed-Quad in an Excribed-Quad

Unread post by tanmoy » Fri Jun 19, 2015 12:49 pm

I have got another solution.I am writing it shortly:
Suppose,$BX$ and $BW$ are perpendiculars to $CI$ and $AI$ respectively and suppose,$DZ$ and $DY$ are perpendiculars to $CI$ and $AI$.
$\measuredangle AID+\measuredangle BIC=0^{0}$.
$\measuredangle BIC+\measuredangle BIX=0^{0}$.
$\therefore \measuredangle BIX=\measuredangle AID$.
$\therefore \measuredangle XBI=\measuredangle XWI=\measuredangle YDI=\measuredangle YZI$.
$\therefore WXYZ$ is cyclic. :mrgreen:
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