2019 IGO Intermediate P2

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Dadgarnia

164 posts

Dadgarnia

#1 h Sep 20, 2019, 9:40 PM • 2 Y

Y by Adventure10, Mango247

Find all quadrilaterals $ABCD$ such that all four triangles $DAB$ , $CDA$ , $BCD$ and $ABC$ are similar to one-another.

Proposed by Morteza Saghafian

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Math-wiz

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Math-wiz

#2 h Sep 21, 2019, 6:25 AM • 1 Y

Y by Adventure10

Was similarity to be considered in that order or different order? If same order, Square. Else, Rectangle

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Dadgarnia

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Dadgarnia

#3 h Sep 21, 2019, 10:44 AM • 1 Y

Y by Adventure10

Math-wiz wrote:

Was similarity to be considered in that order or different order? If same order, Square. Else, Rectangle

Not necessary

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Math-wiz

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Math-wiz

#4 h Sep 21, 2019, 10:52 AM • 1 Y

Y by Adventure10

Dadgarnia wrote:

Math-wiz wrote:

Was similarity to be considered in that order or different order? If same order, Square. Else, Rectangle

Not necessary

Ok, I messed it up. Will get a 0. Meh. Similarity in any order will give all angles equal. Hence all are 90° and it's a rectangle

This post has been edited 1 time. Last edited by Math-wiz, Sep 21, 2019, 10:52 AM

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finava-taklan

20 posts

finava-taklan

#5 h Sep 21, 2019, 11:07 AM • 2 Y

Y by Adventure10, Mango247

Without loss of generality, suppose that ABC is the largest angle of quadrilateral $ABCD$ . We can easily see that 3 angles BCD, CDA, BAD must equal to angle ABC. Therefore, quadrilateral $ABCD$ is a rectangle.

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Pluto1708

1107 posts

Pluto1708

#6 h Sep 21, 2019, 2:01 PM • 3 Y

Y by jrsbr, Adventure10, Mango247

Dadgarnia wrote:

Find all quadrilaterals $ABCD$ such that all four triangles $DAB$ , $CDA$ , $BCD$ and $ABC$ are similar to one-another.

Proposed by Morteza Saghafian

Wait I got square.Toss into the complex plane.
Claim 1 $AC\perp BD$
Proof $DAB\sim CDA\sim BCD\sim ABC$ Hence we see
$\begin{align*} d(d-a)+a(a-c)+b(c-d)=0 \\ c(c-d)+d(d-b)+a(b-c)=0 \\ b(b-c)+c(c-a)+d(a-b)=0 \\ a(a-b)+b(b-d)+c(d-a)=0 \end{align*}$ Adding these we get $a^2+b^2+c^2+d^2=2ac+2bd \implies \left(\dfrac{a-c}{b-d} \right)^2=-1 \implies \left(\dfrac{a-c}{b-d} \right) \in i\mathbb{R} \implies AC\perp BD$ .

Claim 2 $AB\parallel CD$ and $AD\parallel BC$
Proof Now swap the order $DAB\sim BCD\sim CDA\sim ABC$
$\begin{align*} d(c-d)+a(d-b)+b(b-c)=0 \\ b(d-a)+c(a-c)+d(c-d) =0 \\ c(b-c)+d(c-a)+a(a-b) =0 \\ a(a-b)+b(b-d)+c(d-a)=0 \end{align*}$ Adding these we get $2a^2+2b^2-2c^2-2d^2+4cd-4ab=0 \implies \left(\dfrac{a-b}{c-d}\right)^2=1 \implies \left(\dfrac{a-b}{c-d}\right)\in \mathbb{R} \implies AB\parallel CD$ Similarly $AD\parallel BC$

By Claim 1 and Claim 2 we get $ABCD$ a square.

This post has been edited 1 time. Last edited by Pluto1708, Sep 21, 2019, 2:02 PM

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Delta0001

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Delta0001

#8 h Sep 21, 2019, 2:09 PM • 2 Y

Y by Adventure10, Mango247

Pluto1708 wrote:

Solution

From what I can see, you have assumed a constant order of similarity.
What I mean is that your solution assumes that $\Delta DAB \sim \Delta ABC$ however I have a strong feeling that $\Delta DAB \sim \Delta ACB$ could also be true

That would give you answer as square.

However this problem was worded a bit ambiguously that's why this problem arose.

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Dadgarnia

164 posts

Dadgarnia

#9 h Sep 21, 2019, 2:11 PM • 2 Y

Y by Adventure10, Mango247

finava-taklan wrote:

Without loss of generality, suppose that ABC is the largest angle of quadrilateral $ABCD$ . We can easily see that 3 angles BCD, CDA, BAD must equal to angle ABC. Therefore, quadrilateral $ABCD$ is a rectangle.

Your solution just works for convex quadrilateral but $ABCD$ can be concave.

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GeoMetrix

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GeoMetrix

#10 h Sep 21, 2019, 4:00 PM • 1 Y

Y by Adventure10

yup considering that answer is a rectangle.

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InkretBear

85 posts

InkretBear

#11 h Sep 21, 2019, 4:16 PM • 1 Y

Y by Adventure10

The official solution is just rectangles. There are different cases to consider depending on the orientation of the similarities.

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Mathotsav

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Mathotsav

#12 h Sep 21, 2019, 4:52 PM • 1 Y

Y by Adventure10

Wait, official solutions are available? Can anyone share the link?

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InkretBear

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InkretBear

#13 h Sep 21, 2019, 4:53 PM • 1 Y

Y by Adventure10

It's a simple google search

https://igo-official.ir/events/

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Math-wiz

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Math-wiz

#14 h Sep 21, 2019, 5:52 PM • 2 Y

Y by Adventure10, Mango247

InkretBear wrote:

It's a simple google search

https://igo-official.ir/events/

It's in Persian. The file. How to convert it to English

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InkretBear

85 posts

InkretBear

#15 h Sep 21, 2019, 7:05 PM • 1 Y

Y by Adventure10

Change the site to be english first

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Navansh

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Navansh

#16 h Sep 21, 2019, 7:12 PM • 1 Y

Y by Adventure10

But the download is still in the same language

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InkretBear

85 posts

InkretBear

#17 h Sep 22, 2019, 5:20 AM • 2 Y

Y by Adventure10, Mango247

For me it's english, I don't know why it isn't for you.

There I drag and dropped it:
https://igo-official.ir/media/IGO_2019_Booklet_en.pdf

This post has been edited 2 times. Last edited by InkretBear, Sep 22, 2019, 5:21 AM

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Delta0001

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Delta0001

#18 h Sep 22, 2019, 6:20 AM • 2 Y

Y by Adventure10, Mango247

Mathotsav wrote:

Wait, official solutions are available? Can anyone share the link?

https://drive.google.com/file/d/19VL3xM0-uiTMsN5EAoFAdDvA2rQZECNA/view?usp=sharing

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dwip_neel

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dwip_neel

#19 h Oct 10, 2020, 3:14 PM

Y by

Well, very easy as a P2.

Solution:
As $DAB \sim CDA \sim BCD \sim ABC$ ,
$\square ABCD \text{ is cyclic.}$
Let the intersection of diagonals be $O$ .
As we are done up with cyclicity,
$\triangle ABO \sim \triangle AOD$
So, $\angle AOB = \angle AOD = 90^\circ$

So, $ABCD$ is a rhombus. On the other hand, it is cyclic too. So, $ABCD$ must be a rectangle.

Oh, that cannot be concave. No concave exists which satisfies $DAB \sim CDA \sim BCD \sim ABC$ similar triangles.

N.B. How much point for proving it a rhombus?

This post has been edited 3 times. Last edited by dwip_neel, Oct 10, 2020, 3:19 PM

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geometry6

304 posts

geometry6

#20 h Oct 26, 2020, 6:23 PM

Y by

IGO 2019 Intermediate P2 wrote:

Find all quadrilaterals $ABCD$ such that all four triangles $DAB$ , $CDA$ , $BCD$ and $ABC$ are similar to one-another.

Proposed by Morteza Saghafian

Solution. WLOG let $ABCD$ be convex quadrilateral because no concave quadrilateral exists that satisfies the similarity condition. $\triangle DAB \sim \triangle CDA \sim \triangle BCD \sim \triangle ABC \implies ABCD$ is a cyclic quadrilateral. Let diagonals intersect at $P$ , $\triangle ABP \sim \triangle APD$ . $\angle APB = \angle APD = 90^\circ \implies ABCD$ is a rhombus and cyclic $\implies ABCD$ is rectangle. $\blacksquare$

This post has been edited 2 times. Last edited by geometry6, Oct 26, 2020, 6:25 PM

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#22 h Dec 29, 2021, 10:18 AM

Y by

easy to find out that no angle can be greater than the others so answer is all the rectangles.

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