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Source: Sharygin Geometry Olympiad 2012 - Problem 1

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724 posts

nsun48

#1 h Apr 28, 2012, 2:42 AM • 4 Y

Y by ahmedosama, Adventure10, Mango247, and 1 other user

In triangle $ABC$ point $M$ is the midpoint of side $AB$ , and point $D$ is the foot of altitude $CD$ . Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$ .

This post has been edited 4 times. Last edited by nsun48, Apr 28, 2012, 3:21 PM

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1514 posts

ksun48

#2 h Apr 28, 2012, 3:19 AM • 5 Y

Y by Adventure10, Mango247, and 3 other users

If: Let $P$ be the point on $BC$ such that $PA = PB$ . Then $AC = 2DM = \frac{2\cdot CP \cdot MB}{PB} = \frac{CP\cdot AB}{BP}$ , so $\frac{AC}{AP} = \frac{AB}{BP}$ . Then by converse of the Angle Bisector Theorem, $\angle CAP = \angle BAP$ , so $\angle A = 2\angle B$ .

Only If: Let $P$ be the point on $BC$ such that $PA = PB$ . Then $\angle PAB = \angle PBA$ , so $\angle CBA = \angle CAP$ , and triangles $\triangle CAP$ and $\triangle CBA$ are similar. Then $\frac{AC}{AB} = \frac{CP}{PA}=\frac{CP}{PB}=\frac{DM}{MB}=\frac{2DM}{AB}$ , so $AC=2MD$ .

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4394 posts

#3 h Apr 28, 2012, 4:16 AM • 5 Y

Y by Adventure10, Mango247, and 3 other users

Nice idea, nsun48!

The solution could be even faster, seeing that $MP\parallel CD$ .
Then $\hat A=2\hat B\implies\frac{AC}{AB}=\frac{CP}{BP}=\frac{DM}{MB}$ ; as $AB=2BM$ , we get $AC=2DM$ .
Conversely, $AC=2DM\implies AP$ angle bisector of $\angle BAC$ and, with $AP=PB\iff \hat B=\widehat{BAP}$ , we are done! Congratulations!

My idea was based on the relation $BC^2=AC(AB+AC)\iff \hat A=2\hat B$ , then working it with Carnot lemma.

Best regards,
sunken rock

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AndreiAndronache

88 posts

AndreiAndronache

#4 h May 1, 2012, 10:28 AM • 2 Y

Y by Adventure10, Mango247

This is my solution for this problem.

Attachments:

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56 posts

#5 h May 1, 2012, 8:37 PM • 2 Y

Y by Adventure10, Mango247

Denote by $M'D'$ the image of $MD$ in homothety $\mathcal{H}(H,2)$ . It is well-known that $M'D'$ is a chord on the circumcircle $\omega$ of $\triangle ABC$ and we also have $M'D' = 2MD$ . It suffices to note that the inscribed angles corresponding to $AC$ and $M'D'$ are $\angle B$ and $|\angle A - \angle B|$ , respectively. Done.

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315 posts

MBGO

#6 h May 10, 2012, 8:45 AM • 5 Y

Y by DownWithIsrael, Adventure10, Mango247, and 2 other users

Dear all

here is the trigonometric way for this question

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#7 h May 10, 2012, 9:01 AM • 2 Y

Y by Adventure10, Mango247

good job MahanBabol.

and how about the other side?

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315 posts

MBGO

#8 h May 10, 2012, 9:09 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

just put $\measuredangle A=\measuredangle (B+\alpha )$
after some calculating you'll get

$cos\measuredangle A=\frac{sin\measuredangle A.cos\measuredangle B }{sin \measuredangle B } - 1 \equiv 1-sin\measuredangle B.sin\measuredangle \alpha=\frac{cos^{2}\measuredangle B.sin\alpha }{sin\measuredangle B} \equiv \frac{1}{sin\alpha }=\frac{cos^{2}\measuredangle B + sin^{2}\measuredangle B }{sin\measuredangle B }$

which means : $\alpha = \measuredangle B$

And this is the other side Mr. Down With Israel.

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43 posts

#9 h Jul 12, 2012, 5:23 PM • 1 Y

Y by Adventure10

We can use angle chasing:
Denote $N$ is the midpoint of $AC$ , so we have $(DN,AB) \equiv (AB,AN) (mod \pi)$
$(AB,AC) \equiv 2(BC,BA) \Leftrightarrow (DN,AB) \equiv 2(MN,AB) (mod \pi)$ $\Leftrightarrow (DN,MN) + (MN,AB) \equiv (MN,AB) (mod \pi)$ $\Leftrightarrow(DN,MN) \equiv (MN,DM) (mod \pi) \Leftrightarrow DM=DN \Leftrightarrow DM = \dfrac{1}{2}AC$

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21 posts

#10 h Jan 27, 2013, 5:15 PM • 2 Y

Y by Adventure10, Mango247

nsun48 wrote:

In triangle $ABC$ point $M$ is the midpoint of side $AB$ , and point $D$ is the foot of altitude $CD$ . Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$ .

Let $N$ is the midpoint of side $AC$
Thus triangle $DMN$ is isosceles at D
Thus It easily follows that to prove

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21 posts

#11 h Jan 27, 2013, 5:17 PM • 2 Y

Y by Adventure10, Mango247

duanKHTN wrote:

nsun48 wrote:

In triangle $ABC$ point $M$ is the midpoint of side $AB$ , and point $D$ is the foot of altitude $CD$ . Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$ .

Let $N$ is the midpoint of side $AC$
Thus triangle $DMN$ is isosceles at D
Thus It easily follows that to prove

duanKHTN wrote:

nsun48 wrote:

In triangle $ABC$ point $M$ is the midpoint of side $AB$ , and point $D$ is the foot of altitude $CD$ . Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$ .

Let $N$ is the midpoint of side $AC$
Thus triangle $DMN$ is isosceles at D
Thus It easily follows that to prove

nsun48 wrote:

In triangle $ABC$ point $M$ is the midpoint of side $AB$ , and point $D$ is the foot of altitude $CD$ . Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$ .

Let $N$ is the midpoint of side $AC$
Thus triangle $DMN$ is isosceles at D
Thus It easily follows that to prove

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607 posts

#12 h Aug 23, 2013, 2:48 AM • 3 Y

Y by Ankoganit, Adventure10, Mango247

Let $D'$ is the midpoint of $BC$ ,so $MD'||AC$ and $AC=2MD'$ ,now let $MD'=MD$ and ${\angle}A=2X$ ( $X$ is real),so ${\angle}M=2X$ ,hence ${\angle}MDD'=X$ ,but $DD'=BD'$ ,(angle $D$ is right and $D'$ is the midpoint),so ${\angle}ABC=X$ ,hence we are done.

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1916 posts

#13 h Aug 8, 2019, 9:11 PM • 3 Y

Y by Path_to_Almighty, Adventure10, Mango247

Let $\angle CBA=\theta\implies\angle CAB=2\theta$ . Extend $BA$ to a point $K$ such that $AC=AK$ . So, $\angle CKA=\theta$ . So, $\triangle CBD\cong\triangle CKD$ . So, $BD=\frac{BK}{2}=\frac{b+c}{2}$ . So, $DM=\frac{b}{2}$ and for the converse, extend $BD$ to a point $N$ such that $BD=DN$ . We know $DM=\frac{b}{2}$ and $BM=\frac{c}{2}\implies DN=\frac{b+c}{2}$ . So, $AN=(b+c)-c=b=AC\implies \angle NCA=\theta\implies \angle CAB=2\theta$ .

This post has been edited 3 times. Last edited by amar_04, Aug 8, 2019, 9:21 PM

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1587 posts

#14 h Aug 9, 2019, 5:52 AM • 1 Y

Y by Adventure10

It is well known that, $\angle A = 2\angle B \Leftrightarrow {a^2} - {b^2} = bc \Leftrightarrow 2c(MD) = bc.$ Done!

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