Geometry Problem

For unknown reason my other thread was deleted. Probably someone got mad and deleted it with an excuse of going too off topic, I don't know, just guessing. I promised to post another geometry problem with corrections, so here it is. Only people who would like to give it a try are welcome here, for exmaple, Roma, Crawly, Wrum, anaesthetic or anyone else who tried to solve it, rest can back off because your complaints, about how meaningful my thread is, lead to its deletion.

Please prove your answer and don't spit a random guess.

Have fun!

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1. In order to satisfy the condition: GH = HI = HtoAC, one simply draws a circle with centre H (see red and blue diagrams). This is because any contact point at the edge of the circle will be equidistant to the centre of the circle.

2. The AC line can be drawn almost anywhere, so it is no problem to have it line up with G-H-F. (see red diagram)

3. From there, it is obvious that angle x is dependent on the radius of the circle around H.



So once again, there is not a single solution! :p

No man.

Like I said, as long as the lines touch the circle around H, HI = GH = HtoAC is satisfied.

I have mistaken there, so it was deleted before your reply.

Look closely again, because you can prove, that triangles AGF and HFI are exactly the same, it means it defines where C is.

RedSphinx wrote:I have mistaken there, so it was deleted before your reply.

Look closely again, because you can prove, that triangles AGF and HFI are exactly the same, it means it defines where C is.

I looks like a right angle on your diagram. But that is why I had to redraw most of it, because the diagram was giving assumptions that were not stated to be true.

AGF is only equal to HFI if the line F-B crosses I, which it doesn't have to.

And anyway my proof looks pretty convincing to me. Can you disprove it?

cmon man i want my prize! D:

Hmm, let me see. Your screenshot is not very clear. If I find you clear, you state that AC line can be moved within GF and it cannot be defined?

If you draw that circle different radius and you move C on GF, then you go against fact that GH = HI, because GH doesn't change its length and HI does. (Blue and Green examples)

If you put H circle higher (Because it must do 90 degree angle with AB) and not in current center of H then the line DC actually goes through the given circle and it no longer touch it, it goes through it and if given the line which touch the given circle if you draw from same point D, then the line itself would go through the H circle at two points and dont touch at one point. If you draw a line which would touch both circles, then A and C would not meet at same point because AI goes through H, C would go away in distance and A would highly steep and approach G. And A must meet C by touching circles, not going through as well as GH = HI must be satisfied. (Yellow example)

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I actually can't really understand what you said but I see my mistake. I forgot that AI went through H. This fixes A and therefore AC.

Anyway the proof then:

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so x=10

Can you prove that the angle you marked with 60 is 60 by typing? I couldn't see the prove for it to be 60 just by looking at the image.

Like you said,

the triangles AGH and CHI share 1 angle at H, both have a right angle, therefore fixing their 3rd angles.

GH = HI --> AGH and CHI are identical

New diagram:
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CHJ = CHI = AHJ = AGH = GHK = HIK

So yeah. Equilateral triangle --> 180/30 --> 60 degrees


Nice challenge. Wasted so much times on it. |;

It somehow lacks a bit, because you didn't prove that triangle AGH = triangle GHK. I wanted to see you typnig something like:

CHJ = CHI = AHJ = AGH
Angle-AHG = Angle-AHJ = Angle-JHC
Meaning that you get Angle-AHG by this calculation: 180/3 = 60
So Angle-AHC equal 120
So this leads to: 120 + 90 + 90 + Angle-GKI = 360
Angle-HKI = 60

Anyway, you have solved it and I am glad you liked it! It seemed you have seen it through very easily, but if you found if challenging, I'm happy about it. I cant give anything like Gladiator points as I am not allowed to, so perhaps I'll have the honor to give you first rep!

actually i'm unconvinced lol. trying to generalise it but failing.

You aren'? Hmm.
Do you agree that trianges are equal like this CHJ = CHI = AHJ = AGH? It is because that is why you have drawn a circle around H, to prove that each four triangles are exactly the same, because it all have one angle of 90 and two same length lines, right?
So this means that angle AHC is made of three angles of AHG, right? And to make a straight line it takes 180 degrees, so you get like:

y = Angle AHG
3y = 180
y = 60

Now you look at this figure - GHIK. Angle GHI equal 120 as it was proven up there, the figure also contain of two angles of 90, right? So the last angle is GKI, the K angle. So you get this: 120 + 90 + 90 + Angle-GKI = 360

So K angle is 60 degrees.

Ah never mind. I was trying to understand why.

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Basically in this problem the black circle and the line E-F and the 70 degrees angle are almost irrelevant because the real problem is the angle K.

There are many triangles you can draw around the imaginary circle (see diagram), but only an equilateral one will satisfy the conditions.

The vertical line G-C-F marks 1 corner of the triangle. A-I marks a side of the triangle. And A-G-K marks another side of the triangle. I think that's why... Not too sure lolool

Yes, you are right. The trick here was to reveal one of the angles, one of those who can be proved to be the K angle too. If you find an angle which is proven to be K, then you can prove that it is the K. I just picked different way to find K once it was revealed that angle AHG was 60. You can also go this way: Look at triangle GCO, since it is proven that angle GAH is equal to angle HCI due to angle HIC which is 90 as well as another triangle, you can state that an angle GKI is equal to angle GHA. Another way to prove the same though, there are many ways, that's why I admire geometry.

You need the 70 angle corner, because without it you cannot find the X because the triangle is not difined, it can be anything if not the 70 angle corner, try removing the 70 and find X, I couldn't find a way, if you could, great!

geometry is lame.

gif me some algebra or calculus pls

jommetry best

ok maybe not