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We have a regular triangle $ABC$, where side is $15$, point $N$ belongs to $AB$ such that $AN = 5$. Point $M$ belongs to $AC$ such that $AM = 3$. Prove that $BM$ is perpendicular to $CN$.

I tried to make similar triangles carrying the $BK$ segment, where $K$ is the midpoint of $MC$. So, as $NB:AN = MK:AM$ and angle $A$ is common, $ANM$ and $ABK$ triangles are similar, right? Follows that $BK \| NM$.

I have multiple versions of solving, e.g. if we prove that $MK=KC=OK$, where $O$ is the intersection point of $BM$ and $CN$, then $MOC$ triangle will become right, or we can show that $COM$ and $BMH$, where $H$ is the base of height from $B$, triangles are similar.

So, I can't conclude what to do.

dan_fulea
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Zhenya Karapetyan
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  • Looks like a straightforward calculation using law os cosines. First get lengths of BM and CN. Then angles of B at BMC and C at BNC. Third angle you want falls out. – herb steinberg Dec 29 '22 at 22:45
  • Alternatively you can set a coordinate system with origin in, e.g., $A$. Then it's easy to show that $BM$ has gradient $-\frac{\sqrt 3}{9}$, and that line $CN$ has gradient $3 \sqrt 3$, from which you have you thesis. – dfnu Dec 29 '22 at 23:37
  • Or, without coordinate system, after letting $H$ and $K$ be the projections of $M$ and $C$ on $AB$, show that $\triangle MHB \sim \triangle CNK$ and reach the thesis from there. – dfnu Dec 29 '22 at 23:45
  • I thought about setting coordinate system, but the problem is an 8th grade problem, where they have just passed similar triangles. So can we try to solve as simple as possible? Thanks a lot)) – Zhenya Karapetyan Dec 30 '22 at 20:54
  • Is it ok for the 8.th grade to apply the generalized version of Pythagoras e.g. in $\Delta ABM$ to compute the length of $BM$? Similarly then $CN$. To finish, consider the point $S$ on the line $AMC$ between $A$ and $M$ with $AS=1$. So $NS| MB$, $NB$ is a third of $BM$ (or compute this length directly in $\Delta ANS$ as $\sqrt{5^2+1^2-1\cdot 5}$) and in $\Delta SNC$ we know all three sides, so we can check if it has a right angle in $N$ (Pythagoras). If this is not elementary enough, i can drop an answer with a more elementary solution... – dan_fulea Dec 30 '22 at 22:57
  • Thanks for solution. Maybe SN is a third of BM, not NB? – Zhenya Karapetyan Dec 31 '22 at 16:22

2 Answers2

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Here is a solution which arguably can be considered elementary. Let $\Omega$ be the intersection $BM\cap CN$, and draw the cevian $AL$ through $\Omega$ from $A$, $L$ being its intersection with $BC$.

application of the theorem of Ceva

Then $BL=5$ and $LC=10$, since for this placement of $L$ with $BL:LC=1:2$ we have the relation (Ceva) $\displaystyle -1= \frac{NA}{NB}\cdot \frac{LB}{LC}\cdot \frac{MC}{MA}$, which is explicitly forgetting about signs (that match) $\displaystyle -1= \frac{ 5}{10}\cdot \frac{ 5}{10}\cdot \frac{12}{ 3}$.

Draw from $L$ parallels $LM'\|BM$ and $LN'\|CN$ with $M'\in AC$, $N'\in AB$. Then project $N',A,M'$ on $BC$, obtaining the points $N'',A'',M''$.

Using similarities we can compute all lengths as marked in the picture.

Moreover, $CM''$ takes from $CA''=\frac{15}2$ a proportion, which is the same one as $CM':CA=8:15$, so $CM''=\frac 82=4$. From here $LM''=10-4=6$. Also $M'M'':AA''= CM':CA=8:15$ and $AA''=\frac{15\sqrt3}2$ is giving $M'M''=\frac{8\sqrt3}2=4\sqrt 3$.

Doing the same on the other side we get $BN''=\frac{BN'}{BA}BA''=\frac{10/3}{15}\cdot\frac{15}2=\frac 53$, so $N''L=5-\frac 53=\frac{10}3$. And $N'N''=\frac{BN'}{BA}AA''=\frac{10/3}{15}\cdot\frac{15\sqrt 3}2=\frac {5\sqrt 3}3$, so $N''L=5-\frac 53=\frac{10}3$.

It remains to check the similarity $\Delta N''LN'\sim\Delta M''M'L$, the triangles have each a right angle, and we check the proportion: $$ \frac{N''L}{N''N'}= \frac{10/3}{5\sqrt 3/3}= \frac 2{\sqrt 3}= \frac{4\sqrt 3}{6}= \frac{M''M'}{M''L} \ . $$ From here $90^\circ=\widehat{N'LM'}=\widehat{N\Omega M}$.

$\square$

dan_fulea
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  • Why is BL also 5?) Maybe because $$\triangle{CNA}$$ = $$\triangle{BAL}$$, but how? – Zhenya Karapetyan Dec 31 '22 at 19:48
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    This is because of the theorem of Ceva, the point is such that$$-1 = \frac{NA}{NB}\cdot\frac{LB}{LC}\cdot\frac{MC}{MA}\ .$$Two fractions from three are known, the third one determines the position of $L$ on the side $BC$. See also the reference mentioned in the answer, linking to the wiki Ceva. It is a good convention to also use signs, sometimes this convention is irritating, but ok, here is how it works. In our case all points $L,M,N$ are in the interior of the sides. The fraction $NA:NB=-5:10$ is considered with the minus sign, since on the line $AB$ the segments have different directions. – dan_fulea Dec 31 '22 at 19:56
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Expanding a little bit on my comment, let $H$ and $K$ be the projections of $M$ and $C$ on $AB$. Let $O = CN \cap BM$ and $P = CK\cap BM$.

enter image description here

  1. Compute $\overline{MH} = \frac{3\sqrt 3}2$, and $\overline{HB} = 15-\frac32 = \frac{27}2$.
  2. Compute $\overline{CK} = \frac{15\sqrt 3}2$, and $\overline{NK} = \frac{15}2-5 = \frac52$.
  3. Observe that $$\frac{\overline{HB}}{\overline{MH}} = \frac{\overline{CK}}{\overline{NK}} = 3\sqrt3,$$hence $\triangle MHB \sim\triangle CNK$ by SAS criterion. In particular $\angle NCK \cong \angle MBH$.
  4. By 3. and the fact that $\angle KPB \cong \angle CPO$ (vertical angles) we get that $\triangle COP \sim \triangle PKB$ by AA criterion. Hence the thesis.
dfnu
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