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

Math-wiz wrote: Was similarity to be considered in that order or different order? If same order, Square. Else, Rectangle Not necessary

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

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.

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.

Pluto1708 wrote:

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.

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.

yup considering that answer is a rectangle.

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

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

It's a simple google search https://igo-official.ir/events/

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

Change the site to be english first

But the download is still in the same language

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

Mathotsav wrote: Wait, official solutions are available? Can anyone share the link? https://drive.google.com/file/d/19VL3xM0-uiTMsN5EAoFAdDvA2rQZECNA/view?usp=sharing

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?

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.$\blacksquare$

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