BdMO National Higher Secondary 2020 P5

[Mursalin]

ত্রিভুজ \(ABC\)-এ \(AB=52, BC=34\) আর \(CA=50\)। আমরা \(BC\) বাহুর মাঝে \(n-1\) সংখ্যক বিন্দু এমন ভাবে নিই যাতে \(BC\) রেখাংশ \(n\)টা সমান ভাগে বিভক্ত হয়। এই বিন্দুগুলোর মধ্যে যদি \(A\) থেকে \(BC\)-এর ওপর আঁকা লম্বের পাদবিন্দু, \(A\) থেকে \(BC\)-এর ওপর আঁকা মধ্যমার পাদবিন্দু আর \(A\) কোণের সমদ্বিখণ্ডকের পাদবিন্দু থাকে, তাহলে \(n\)-এর সম্ভাব্য সর্বনিম্ন মান কত?


In \(\triangle ABC\), \(AB = 52, BC = 34\) and \(CA = 50.\) We split \(BC\) into \(n\) equal segments by placing \(n-1\) new points. Among these points are the feet of the altitude, median and angle bisector from \(A\). What is the smallest possible value of \(n\)?

[Mehrab4226]

We, claim that $min(n) = 102$
Now the median divides BC into $17-17$ segments. And the Altitude divides BC into $20-14$ segments(By Pythagoras). And the angle bisector divides the same in $\frac{52}{3}-\frac{50}{3}$(Using the theorem of the ratio of the sides equal ratio of the corresponding segments made by the angle bisector).

If they were all integers the GCD of them would have been the maximum length of each segment. So now we need a number that perfectly divides $1$(GCD of 17,20,14) , $\frac{52}{3}, and \frac{50}{3}$. There are infinite numbers with the property but we need the max of them.
Let the number be $g$ then,
$1 = kg$ for some integer k
$\frac{52}{3}= tg$ for some integer t
$\frac{\frac{52}{3}}{1} = \frac{g}{t}$
$t : k = 52 : 3$
To get the maximum of g we need the minimum of t and k that is 52 and 3 respectively.
Thus $g = \frac{1}{3}$

Now the max length of each segment $= \frac{1}{3}$
The minimum value of $n= \frac{34}{\frac{1}{3}} = 102$ (Ans)