Mock AIME I 2012 Problems/Problem 3

Revision as of 17:31, 7 April 2012 by Esque (talk | contribs) (Created page with "==Problem== Triangle <math>MNO</math> has <math>MN=11</math>, <math>NO=21</math>, <math>MO=23</math>. The trisection points of <math>MO</math> are <math>E</math> and <math>F</mat...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Triangle $MNO$ has $MN=11$, $NO=21$, $MO=23$. The trisection points of $MO$ are $E$ and $F$, with $ME<MF$. Segments $NE$ and $NF$ are extended to points $E'$ and $F'$ such that $ME' || NF$ and $OF' || NE$. The area of pentagon $MNOF'E'$ is $p\sqrt{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Use Heron's formula to find $A=[MNO]=\frac{33}{4}\sqrt{195}$. Also note from the trisection that $[NME]=[NEF]=[NFO]=A/3$. Now $ME'\parallel NF,ME=EF\Longrightarrow NE=EE'$. Similarly, $NF=FF'$. From this we deduce that

(i) $[NEM]=[MEE']\Longrightarrow [NME']=2[NME]=\frac{2}{3}A$

(ii) Similarly to (i), $[NOF']=\frac{2}{3}A$

(iii) $\triangle NEF\sim\triangle NE'F'$ with ratio $\frac{1}{2}$, so $[NE'F']=4[NEF]=\frac{4}{3}A$

The area of our pentagon is $[NME']+[NOF']+[NE'F']=\frac{2}{3}A+\frac{2}{3}A+\frac{4}{3}A=\frac{8}{3}\cdot \frac{33}{4}\sqrt{195}=22\sqrt{195}$. The answer is $22+195=\fbox{217}$.

Invalid username
Login to AoPS