BdMO National Secondary 2020 P4

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BdMO National Secondary 2020 P4

Unread post by Mursalin » Wed Feb 03, 2021 10:19 pm

\(ABCD\) একটি বর্গক্ষেত্র। \(P\) এবং \(Q\) যথাক্রমে \(BC\) এবং \(CD\) রেখাংশের উপর দুইটি বিন্দু যাতে করে \(\angle APQ = 90^\circ\) হয়। দেওয়া আছে যে, \(AP = 4\) এবং \(PQ= 1\)। যদি \(AB\)-এর দৈর্ঘ্যকে লঘিষ্ঠ আকারে \(\frac{m}{n}\) হিসেবে লেখা যায়, তবে \(m+10n\)-এর মান বের কর।

\(ABCD\) is a square. \(P\) and \(Q\) are two points in segment \(BC\) and \(CD\) respectively such that \(\angle APQ = 90^\circ\). It is given that \(AP = 4\) and \(PQ = 1\). If we express the length of segment \(AB\) as \(\frac{m}{n}\) in lowest terms, compute \(m+10n\).
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Re: BdMO National Secondary 2020 P4

Unread post by Mehrab4226 » Sat Feb 06, 2021 12:03 am

This problem can be solved by making three equations with three variables. But I am still looking for an easier solution.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Joined:Sun Jan 17, 2021 7:51 pm

Re: BdMO National Secondary 2020 P4

Unread post by Ridwan056 » Fri Feb 12, 2021 10:32 am

I think we can use similar triangles. Triangle APB and triangle PCQ are similar. Hence, AB= 4CP, which implies BP=3CP; and BP=4CQ. so, AP^2= 16= AB^2 +BP^2. Similarly, applying like this on PCQ gives BP=12/5. Therefore, AB=16/5. The final answer is m+10n= 16+50=66.

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