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Let $ABC$ be a triangle with $AB = 1$, $AC = 2$ and $m(\widehat{BAC}) = 30^\circ$. We build on the outside the equilateral triangles $ABM$ and $ACN$.

Let $D$, $E$ and $F$ be the midpoints of $AM$, $AN$ and $BC$.

a) Prove that triangle $DEF$ is equilateral.

b) Find $x$ and $y$ so that $\overrightarrow{MN} = x\overrightarrow{AB} + y\overrightarrow{AC}.$

For a) all I found is that $DE = \frac{\sqrt{5 + 2\sqrt{3}}}{2}$. For b) I have no idea how I should start.

Thanks in advance!

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  • $\begingroup$ Are you sure it is not $\angle ACB=30°$? $\endgroup$ – Aretino Jul 17 '16 at 18:09
  • $\begingroup$ @Aretino The problem clearly says it's about $\angle BAC$. $\endgroup$ – George R. Jul 17 '16 at 18:12
  • $\begingroup$ @GeorgeR. D you realize that triangle ABC is in fact a 30-60-90 triangle? AC =2 is the hypotenuse, right angle is at B $\endgroup$ – imranfat Jul 17 '16 at 18:30
  • $\begingroup$ @imranfat sine rule doesn't valid in your case $\endgroup$ – Aman Rajput Jul 17 '16 at 18:49
  • $\begingroup$ I don't know what is wrong, I've checked the problem over and over and I copied it as it is on the paper. $\endgroup$ – George R. Jul 17 '16 at 18:49
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enter image description hereLet G be the midpoint of AC.

By midpoint theorem and some angle chasing, we have $\triangle DAE \cong \triangle FGE$.

Then, ED = EF and $\angle DEF = 60^0$ (from the fact that $\alpha = \alpha’$ and using $\angle DEG$ as the middle man).

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  • $\begingroup$ Nice! Your proof does not rely on $\angle BAC=60°$ or $AC=2AB$, which confirms that this is a very general result. $\endgroup$ – Aretino Jul 18 '16 at 13:51
  • $\begingroup$ @Aretino I think you are right even if $\angle BAC$ is obtuse. $\endgroup$ – Mick Jul 18 '16 at 16:07
  • $\begingroup$ This figure is made by which software? $\endgroup$ – Aman Rajput Jul 19 '16 at 12:05
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    $\begingroup$ @AmanRajput Geogebra then PC Paint. $\endgroup$ – Mick Jul 19 '16 at 15:06
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This is very easy using coordinates. Notice that $AB\perp AN$, so we can set $A=(0,0)$, $B=(0,1)$, $N=(2,0)$, $M=(-\sqrt3/2,1/2)$, $C=(1,\sqrt3)$.

It is then straightforward to find the coordinates of midpoints $D$, $E$, $F$ and check that $DE=EF=FD$, as well as finding $x$ and $y$ such that $\overrightarrow{MN} = x\overrightarrow{AB} + y\overrightarrow{AC}$.

It would be nice to prove that $DEF$ is equilateral by a purely geometrical argument. I haven't succeeded yet, but I noticed that this result has a more general validity: it is true for any value of $\angle BAC$, and for any length of $AB$ and $AC$.

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  • $\begingroup$ I've added a geometrical proof. $\endgroup$ – Mick Jul 18 '16 at 13:35

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