Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

3. As shown in Figure 1, in △ABC\triangle A B C, OO is the midpoint of side BCB C, and a line through OO intersects lines ABA B and ACA C at two distinct points MM and NN respectively. If
AB→=mAM→,AC→=nAN→, \begin{array}{l} \overrightarrow{A B}=m \overrightarrow{A M}, \\ \overrightarrow{A C}=n \overrightarrow{A N}, \end{array}

then m+n=m+n=

A number or a short expression. Spacing and $ signs are ignored.

Solution

3. 2 .

Solution 1 Notice that,
AO→=12(AB→+AC→)=m2AM→+n2AN→. \overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} .

Since points M,O,NM, O, N are collinear, we have,
m2+n2=1⇒m+n=2.  \frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }

Solution 2 Since points M,O,NM, O, N are on the sides or their extensions of △ABC\triangle A B C, and the three points are collinear, by Menelaus' theorem we get
AMMB⋅BOOC⋅CNNA=1. Hence n−11−m=1⇒m+n=2. \begin{array}{l} \frac{A M}{M B} \cdot \frac{B O}{O C} \cdot \frac{C N}{N A}=1 . \\ \text { Hence } \frac{n-1}{1-m}=1 \Rightarrow m+n=2 . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.