regional mo 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.
kh ibrahim
Posts: 17
Joined: Mon May 09, 2016 11:18 am

regional mo 2015

Unread post by kh ibrahim » Sat Dec 10, 2016 12:49 pm

this is a problem of regional math olympiad 2015
Attachments
rps20161210_124814.jpg
rmo 2016
rps20161210_124814.jpg (57.18 KiB) Viewed 1746 times

kh ibrahim
Posts: 17
Joined: Mon May 09, 2016 11:18 am

Re: regional mo 2015

Unread post by kh ibrahim » Mon Dec 12, 2016 11:40 pm

Disclose the first approach

User avatar
asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

Re: regional mo 2015

Unread post by asif e elahi » Tue Dec 13, 2016 12:53 am

Use the Power of Point theorem to prove that $PA\times PD=PE^2=PB\times PC$

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: regional mo 2015

Unread post by Absur Khan Siam » Mon Jan 30, 2017 11:17 pm

asif e elahi wrote:Use the Power of Point theorem to prove that $PA\times PD=PE^2=PB\times PC$
$PA \times PD = 12 \times (PA + AB + BC + CD) = 12 \times (12 + AB + BC + 2AB)$
$ = 36AB + 12BC + 144...(i)$
$PB \times PC = (PA+AB) \times (PA + AB + BC) = (12+AB) \times (12 + AB + BC)$
$ = AB^2 + 24AB + 12AB + AB \times BC ...(ii)$
$(i) = (ii) \rightarrow 36AB + 12BC + 144 = AB^2 + 24AB + 12AB + AB \times BC$
$ \rightarrow 12AB = AB(AB+BC)$
Thus,$AB+BC = 12$
And we get , $PC = PA + AB + BC = 24$ ;)
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Post Reply