Let ABC be a triangle and M be a point on sid | vedclass

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(A) Let the area of $	riangle ABC$ be $S$. Given that $MN \parallel BC$,$	riangle ANM \sim 	riangle ABC$. Let $AM/AC = k$. Then $	ext{Area}(	rian

Vedclass · Accepted Answer

$5$

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(A) Let the area of $	riangle ABC$ be $S$. Given that $MN \parallel BC$,$	riangle ANM \sim 	riangle ABC$. Let $AM/AC = k$. Then $	ext{Area}(	riangle ANM) = k^2 S$. Since $M$ is closer to $C$ than $A$,$AM > MC$,so $k > 1/2$. Since $MP \parallel AB$,$	riangle MPC \sim 	riangle ABC$. Let $MC/AC = 1-k$. Then $	ext{Area}(	riangle MPC) = (1-k)^2 S$. Quadrilateral $BNMP$ is a parallelogram because $MN \parallel BP$ and $MP \parallel NB$. Area$(BNMP)$ = $S - 	ext{Area}(	riangle ANM) - 	ext{Area}(	riangle MPC) = S - k^2 S - (1-k)^2 S = S(1 - k^2 - (1 - 2k + k^2)) = S(2k - 2k^2) = 2k(1-k)S$. Given $2k(1-k)S = \frac{5}{18}S$,we have $2k - 2k^2 = \frac{5}{18}$,or $36k^2 - 36k + 5 = 0$. Solving for $k$: $k = \frac{36 \pm \sqrt{1296 - 720}}{72} = \frac{36 \pm \sqrt{576}}{72} = \frac{36 \pm 24}{72}$. So $k = \frac{60}{72} = \frac{5}{6}$ or $k = \frac{12}{72} = \frac{1}{6}$. Since $AM > MC$,$k > 1/2$,so $k = 5/6$. Then $AM/AC = 5/6$,which implies $MC/AC = 1/6$. Thus,$AM/MC = (5/6) / (1/6) = 5$.

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