secondary dhaka divisional 2015
Forum rules
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
-
- Posts:2
- Joined:Sat Aug 24, 2013 6:46 pm
$ABCD$ is a parallelogram and it's diagonals meet at point $O$. $P$ an $Q$ are the midpoints of $AO$ and $BC$ consecutively. $\angle A = \angle DPQ$ and $\angle DBA = \angle DPQ$. If $AB = 2$ unit, then find out the area of $ABCD$.
Last edited by Phlembac Adib Hasan on Mon Jan 02, 2017 8:04 pm, edited 1 time in total.
Reason: Latexed
Reason: Latexed
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 pm
Re: secondary dhaka divisional 2015
Interestingly, this problem needs spiral similarity.
So, we have, $\triangle DAB$ similar to $ \triangle DPQ $.
By spiral similarity, $\triangle DQB $ similar to $ \triangle DPA $.
So, $\angle DAP=\angle DBQ$. This is only possible when the parallelogram is a rectangle.
We also have, for similarity reasons, $DA/AP = DB/BQ$ or, $DA/AO = DB/BC$.
Along with the condition that, $\angle DAP=\angle DBQ$ we get that $\triangle AOD $ similar to $\triangle BCD$. We get that, $\angle AOD =90$. So, we get, $ABCD$ is a square. So the answer is $4$.
So, we have, $\triangle DAB$ similar to $ \triangle DPQ $.
By spiral similarity, $\triangle DQB $ similar to $ \triangle DPA $.
So, $\angle DAP=\angle DBQ$. This is only possible when the parallelogram is a rectangle.
We also have, for similarity reasons, $DA/AP = DB/BQ$ or, $DA/AO = DB/BC$.
Along with the condition that, $\angle DAP=\angle DBQ$ we get that $\triangle AOD $ similar to $\triangle BCD$. We get that, $\angle AOD =90$. So, we get, $ABCD$ is a square. So the answer is $4$.
Frankly, my dear, I don't give a damn.