Cyclic quadrilateral area(Maths olympiad) | Mathematics | Page 1

queen voidBOT Sep 4, 2024, 1:12 PM

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subtle vigil Sep 4, 2024, 1:22 PM

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Well you can calculate $\cos{135}=\cos{(90+45)}=\cos{90}\cos{45}-\sin{90}\sin{45}$

heady fiberBOT Sep 4, 2024, 1:23 PM

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God of Arctic

austere kelp Sep 4, 2024, 1:35 PM

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We can use

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Area of Cyclic quadrilatral is given by
$$ \sqrt{(s-a)(s-b)(s-c)(s-d)} $$ where s is semi-perimeter and a,b,c,d are sides of cyclic quadrilateral

heady fiberBOT Sep 4, 2024, 1:36 PM

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roltt

austere kelp Sep 4, 2024, 1:37 PM

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( Brahmagupta's Formula )

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Perhaps that could be useful to your question

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Well, all you need to do is cosine law bash.

austere kelp Sep 4, 2024, 1:44 PM

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heady fiber **God of Arctic**

All you need to do is to use this to find cos135 value

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Or even better cos(90+x)=-sin(x)

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so 135=90+45
cos(135)=cos(90+45)=-sin(45)

austere kelp Sep 4, 2024, 2:04 PM

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Find radius and then the sides

austere kelp Sep 4, 2024, 2:04 PM

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heady fiber **roltt**

and use this formula finally

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Here s=a+b+c+d/2

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You got the answer?

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Oh okay.

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Yes

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That was a wrong assumption, I see.

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You can still find QR

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P,O,R aren't collinear though

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so you can't say that QOR=45

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Np

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QSR=QPR=22.5 (chord QR)

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same for PRS=22.5

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talking abt the angles

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so angle SOR=135 degrees

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in SOR you can use cosine law now and find SR

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We don't rlly need to prove them collinear

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Just consider triangle SOR

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you have got 2 angles in SOR so you got the third angle as well

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Well you're right

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Those points arent necessarily collinear

austere kelp Sep 4, 2024, 2:36 PM

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austere kelp QSR=QPR=22.5 (chord QR)

So, this doesn't hold good either.

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Lets use the cyclic properties

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Last thing which we can do

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I've probably found out a way, gimme some more time

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this is a hard problem

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easily 3 marker

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I've got it though

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Are you comfortable with trigo?

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I'm in the same country

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Okay, area of a triangle is 1/2 absin(x) where x is the angle bw the sides a and b

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Yeah

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And you would need to know jensen inequality for this

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Np

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I'm sending a picture of the solution in like 10 mins

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I'll quickly explain that inequality

austere kelp Sep 4, 2024, 2:55 PM

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austere kelp Okay, area of a triangle is 1/2 absin(x) where x is the angle bw the sides a and...

clear with this?

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r=1, you sure?

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nice

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okay

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WhatsApp_Image_2024-09-04_at_8.38.29_PM_1.jpeg

WhatsApp_Image_2024-09-04_at_8.38.29_PM.jpeg

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You can do the calculations

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@austere kelp

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pls ask if you've any doubts

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this is honestly a 5 mark problem

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alright

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Its at least a 3 marker lol

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WhatsApp_Image_2024-09-04_at_8.45.03_PM.jpeg

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Sorry, bad camera, pls co-operate

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Not necessary

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for maximum area, yes

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You will have to consider a sqaure for max area

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so that altitude length is maxmium

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as for why S is the midpoint

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suppose S isn't the midpoint

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now try drawing altitudes from S to PR

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move S from a point, which is close to P to a point close to R

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No it is not

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srry for terrible diagram

austere kelp Sep 4, 2024, 3:31 PM

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austere kelp move S from a point, which is close to P to a point close to R

and draw altitudes

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Now when do you think that the altitude lenth will be max

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Yes

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Right

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So, thats why S is midpoint for maximum area

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yes

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Yes

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Its possible to prove that with trigonometry too

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Hey sorry

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Its not a square

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PQ is not necessarily equal to the other 3 sides

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We cant prove that PQ is equal to the other sides

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yes

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and if it were actually a square then poq should have been 45

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So, yes not a square

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How?

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SR=PS=RQ.....

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opposite sides are equal in rectangle

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?

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not a square, not a rect.

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I dont think so

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so, how confident are you?

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Yes

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huh

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My MT will start from 9th...:(

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same

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Am doing nothing except for maths

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good

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same

subtle vigil Sep 4, 2024, 4:06 PM

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Say thanks @austere kelp

torpid quailBOT Sep 4, 2024, 4:06 PM

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@subtle vigil has given 1 rep to @gleaming jasper @sand mural

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@sand mural has given 1 rep to @gleaming jasper

austere kelp Sep 4, 2024, 4:13 PM

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a=b=y (used a,b,y for alpha,beta,gamma)

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so a+b+y+135=360

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3a=225

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a=75

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Thx.

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@austere kelp

torpid quailBOT Sep 4, 2024, 4:15 PM

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@gleaming jasper has given 1 rep to @sand mural

subtle vigil Sep 4, 2024, 4:16 PM

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austere kelp Thx.

Thanks bruv

torpid quailBOT Sep 4, 2024, 4:16 PM

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@subtle vigil has given 1 rep to @gleaming jasper

subtle vigil Sep 4, 2024, 5:12 PM

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Perhaps because they subtend the arc of equal length (assuming O is the center)

torpid quailBOT Sep 4, 2024, 7:04 PM

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@sand mural has given 1 rep to @subtle vigil

#Cyclic quadrilateral area(Maths olympiad)