For discussing Olympiad level Geometry Problems
sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am

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.

Last edited by sowmitra on Mon May 04, 2015 8:38 pm, edited 1 time in total.
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

- 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.

sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am

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

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

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
Posts: 110
Joined: Wed Mar 20, 2013 10:50 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

whatever, now i got the solution which nirjhor posted

tanmoy
Posts: 305
Joined: Fri Oct 18, 2013 11:56 pm
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.