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In the figure below $DM=DN$ and $\dfrac{Area_{({\Delta ADN})}}{Area_{({\Delta ABC})}}=\dfrac{6}{121}$.

What is $\dfrac{AB}{AD}?$

My attempt:

Knowing that $\mathbf{Area_{({\Delta ADN})}}\mathbf{=}\mathbf{Area_{({\Delta ADM})}}$, and since the distances of $\mathbf{M}$ and $\mathbf{N}$ to $\mathbf{AB}$ are the same, I tried to find a relation between $\mathbf{Area_{({\Delta ABC})}}$ and $\mathbf{Area_{({\Delta ABM})}}$ in order to compare $\mathbf{AB}$ and $\mathbf{AD}$, but I couldn't reach to any good result.

I really appreciate your help.

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    $\begingroup$ What is $S_{ABC}$? $\endgroup$ Commented Feb 19, 2024 at 19:24
  • $\begingroup$ @jjagmath, The area. $\endgroup$ Commented Feb 19, 2024 at 19:40
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    $\begingroup$ Don't just answer in the comments. Edit your question to improve it. People shouldn't have to read comments to understand your question. $\endgroup$ Commented Feb 19, 2024 at 19:47
  • $\begingroup$ Do you know how to relate $ NA / AC$ to $CB / BM$? Or to $ AD / DB$? $\endgroup$ Commented Feb 19, 2024 at 20:11

2 Answers 2

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Draw from $N$ the parallel $EN$ to $CM$ and set: $$ AD=1,\quad BD=DE=x,\quad AE=1-x,\quad BM=EN=h,\quad BC=k. $$ The given ratio of areas can be written as: $$ {h\over(1+x)\,k}={6\over121}, \quad\text{that is:}\quad {h\over k}={6\over121}(1+x). $$ On the other hand, from the similarity of triangles $AEN$ and $ABC$ we get: $$ {h\over k}={1-x\over1+x}. $$ By comparison we thus obtain: $$ {6\over121}(1+x)={1-x\over1+x}, $$ which can be solved to $x=5/6$. Hence: $$ {AB\over AD}=1+x={11\over6}. $$

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    $\begingroup$ Excellent and bravo. $\endgroup$ Commented Feb 19, 2024 at 22:42
  • $\begingroup$ As an aside, without constructing $N$, we could use Menelaus theorem (which is more machinery) to get those ratios. $\endgroup$ Commented Feb 20, 2024 at 4:39
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This is solvable using vector methods. Let

$ u = DM$

$ v = DB$

Let $c \gt 0 $ be such that $AD = c v $

Also let D be the origin. Then

$ M = u $

$ B = v $

$ A = - c v $

$ N = - u $

Now connect $M$ to $B$ and $A$ to $N$, then these two lines have equations

$ \ell_{MB} = M + t (MB) = u + t (v - u) $

$ \ell_{AN} = A + s (AN) = - c v + s (- u + c v ) $

Intersecting, the two lines to find point $C$, we get the following linear system in $t,s$:

$1 - t = - s$

$t = - c + s c $

Thus, $t = - c + c (t - 1) $ from which $t = \dfrac{ 2 c}{c - 1} $

Therefore,

$ C = u + t (v - u) = (1 - t) u + t v = \dfrac{- 1 - c}{c - 1} u + \dfrac{ 2c}{c-1} v$

Now let $[.]$ denote the area, then

$[ADN] = \dfrac{1}{2} c \| v \times u \| $

and

$ [ABC] = \dfrac{1}{2} \| (CB) \times (CA) \| = \dfrac{1}{2} \| \left( \dfrac{c+1}{c-1} u + \dfrac{-c - 1 } { c-1} v \right) \times \left( \dfrac{c+1}{c -1} u + \dfrac{ - c - c^2}{ c-1} v \right) \| \\= \dfrac{1}{2} \left| \dfrac{ (c+1)(- c - c^2) + (c+1)(c+1)} {(c- 1)^2} \right| \| u \times v \| \\ =\dfrac{1}{2} \| u \times v \| \left| \dfrac{ (c+1) (- c^2 + 1 ) }{(c-1)^2} \right| $

The ratio of areas is known, this will determine $c$. Now, we must have $ c \gt 1$ to have the proper relative position of the points as in the attached figure. Therefore,

$ 6 [ABC] = 121 [ADN] $

$ 6 \ (c+1) \ ( c^2 - 1)= 121 \ c ( c - 1)^2 $

So that

$ 6 (c+1)^2 = 121 c (c-1) $

which expands to

$ 6 c^2 + 12 c + 6 = 121 c^2 - 121 c $

And finally,

$ 115 c^2 - 133 c - 6 = 0 $

The positive solution is $ c = \dfrac{6}{5} $

$ \dfrac{AB}{AD} = \dfrac{1 + c}{c} = \dfrac{11}{6} $

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  • $\begingroup$ So for the given area ratio D is the midpoint of AB; OK ? $\endgroup$ Commented Feb 19, 2024 at 21:20

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