BdMO National Higher Secondary 2020 P9

[Mursalin]

\(ABCD\) একটি উত্তল চতুর্ভুজ। \(AC\) এবং \(BD\)-এর ছেদবিন্দু \(O\)। \(AO=3,BO=4,CO=5,DO=6\)। \(X\) এবং \(Y\) যথাক্রমে \(AB\) ও \(CD\) বাহুর উপর অবস্থিত দুইটি বিন্দু যাতে \(X, O, Y\) বিন্দু তিনটি সমরেখ হয়। \(XB/XA+YC/YD\)-এর সর্বনিম্ন মানকে \(\frac{a}{b}\sqrt{c}\) আকারে প্রকাশ করা যায়, যেখানে \(a\), \(b\) পরস্পর সহমৌলিক ধনাত্মক পূর্ণসংখ্যা এবং \(c\), \(1\)-এর চেয়ে বড় কোনো বর্গ সংখ্যা দ্বারা নিঃশেষে বিভাজ্য নয়। \(10a+b+c\)-এর মান কত?


Let \(ABCD\) be a convex quadrilateral. \(O\) is the intersection of \(AC\) and \(BD\). \(AO=3, BO=4, CO=5, DO=6\). \(X\) and \(Y\) are points in segment \(AB\) and \(CD\) respectively such that \(X, O, Y\) are collinear. The minimun of \(XB/XA+YC/YD\) can be written as \(\frac{a}{b}\sqrt{c}\) , where \(\frac{a}{b}\) is in lowest terms and \(c\) is not divisible by any square number greater than \(1\). What is the value of \(10a+b+c\)?

[Mehrab4226]

Let us take any line $XY$ through
Let,
$\angle AOX = \angle COY = \theta_1$
$\angle BOX = \angle DOY = \theta_2$
$\frac{\sin\theta_1}{\sin\theta_2}=x$
$\angle AXO= \alpha$
$\therefore \angle OXB=180-\alpha$
$\angle OYC = \beta$
$\therefore \angle DYO= 180- \beta$

The figure will be helpful.

Screenshot 2021-03-17 23.29.33.png (49.73KiB)Viewed 1699 times

By sine rule on $\triangle AOX$ we get,
$\frac{AX}{\sin \theta_1}=\frac{AO}{\sin \alpha}$
$\therefore AX= \frac{AO \times \sin \theta_1}{\sin \alpha}$
$\therefore AX= \frac{3 \sin \theta_1}{\sin \alpha} \cdots (1)$
Similary in $\triangle BOX$,
$BX=\frac{BO \times \sin \theta_2}{\sin 180-\alpha} = \frac{BO \times \sin \theta_2}{\sin \alpha}$[Since $\sin \alpha = \sin 180-\alpha$]
$\therefore BX=\frac{4 \sin \theta_2}{\sin \alpha} \cdots (2)$
By $(2) \div (1)$,
$\frac{BX}{AX}=\frac{4 \sin \theta_2}{\sin \alpha} \times \frac{\sin \alpha}{3 \sin \theta_1}= \frac{4}{3} \times \frac{\sin \theta_2}{\sin \theta_1} = \frac{4}{3}(\frac{1}{x}) \cdots (I)$

Doing the same thing with $\triangle COY$ and $\triangle DOY$ we get,
$\frac{CY}{DY}=\frac{5}{6}(x) \cdots (II)$

Adding$(I)$ and $(II)$ we get,
$\frac{BX}{AX}+\frac{CY}{DY}=\frac{4}{3x}+\frac{5x}{6}=\frac{8+5x^2}{6x}$

Again by AM-GM we get,
$\frac{8+5x^2}{2} \geq \sqrt{8\times 5x^2}$
Or,$\frac{8+5x^2}{2} \geq 2x\sqrt{10}$
Or,$\frac{8+5x^2}{2x} \geq 2\sqrt{10}$
Or,$\frac{8+5x^2}{6x} \geq \frac{2}{3}\sqrt{10}$\
$\therefore \frac{BX}{AX}+\frac{CY}{DY} \geq \frac{2}{3}\sqrt{10}$

So, $a=2$,$b=3$, and $c=10$
$\therefore 10a+b+c=10 \times 2 +3+10 = 33$(Ans)