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Dhaka regional 16 P4
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Dhaka regional 16 P4
Let the reflection of $X$ across $Y$ be $Z$. Now, see that $\angle PQX=\angle PRZ, \angle PQR=\angle PXY=\angle PZR$ and finally $PQ=PR$. These imply that $\triangle PQX\cong \triangle PRZ$. So $QX=RZ$, and thus $QX+RX=RZ+RX=XZ=2XY=24$
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: Dhaka regional 16 P4
How can you write $XZ = 2XY$?
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 pm
Re: Dhaka regional 16 P4
As $Z$ is defined as the reflected point of $X$ with right to $Y$, we get $ XY=YZ$. From there we get $XZ=2XY$.
Frankly, my dear, I don't give a damn.
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: Dhaka regional 16 P4
Sorry!I missed some questions. How ${\angle}PQX = {\angle}PRZ$ and${\angle}PQR = {\angle}PXY$?
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Dhaka regional 16 P4
Learn the properties of cyclic quads. More than $50\%$ of all olympiad geometry problems exploit the usage of cyclic quad. https://www.expii.com/t/when-is-a-quadr ... on&id=1795Absur Khan Siam wrote:Sorry!I missed some questions. How ${\angle}PQX = {\angle}PRZ$ and${\angle}PQR = {\angle}PXY$?
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.