Challenging geometry problem

Discus: What Are My Chances?: December 2002 Archive: Challenging geometry problem
 By Pat57575 (Pat57575) on Sunday, December 29, 2002 - 12:04 am: Edit

An isosceles triangle has side lengths of 10, 10, and 8. A circle is inscribed in the triangle. Find the area of the circle in terms of pi.

 By caltech07 on Sunday, December 29, 2002 - 12:43 am: Edit

48/7 pi

 By Burberry (Burberry) on Sunday, December 29, 2002 - 12:43 am: Edit

omg, impossible!!! my brain is about to explode

 By caltech07 on Sunday, December 29, 2002 - 12:46 am: Edit

nah. put it on a coordinate plane, use trig, keep identities in mind, no calculator.

 By 800m on Sunday, December 29, 2002 - 12:58 am: Edit

Here is how I solved it:

let r be the radius, O be the center. AB=AC=10, BC=8 then the height AD=84^1/2

and we can have this ratio:AB/BD=AO/r => 10/4=AO/r => AO=5/2r

also AO+r=AD, so r+5r/2=84^1/2=>r^2=48/7, therefore the area is 48Pi/7.

really impressive caltech07!

 By 800m on Sunday, December 29, 2002 - 12:59 am: Edit

but i don't think you need the plane or trig or whatsoever

 By caltech07 on Sunday, December 29, 2002 - 01:01 am: Edit

ah, very nice and more elegant than mine. i always resort to trig for some reason, but your solution is excellent.

 By 800m on Sunday, December 29, 2002 - 01:03 am: Edit

thanks, by the way, congratulations on being a caltech frosh next year!

 By caltech07 on Sunday, December 29, 2002 - 01:12 am: Edit

thanks

 By vp on Sunday, December 29, 2002 - 03:13 am: Edit

ok one sec 800m....u took it that way...
i took the ratio AD/BD = AO/r

why the hell dont i get the same answer......hmmmmm

 By Mitwannabe (Mitwannabe) on Sunday, December 29, 2002 - 04:19 am: Edit

i tried both, not sure why vp's doesn't work, i haven't done identities yet, but i know enuff triangle circle properties to do this one, it's easy, just not sure, i get like (48/7)pi with one, and like 24.3611 with the other, don't ask me for it in pi, haha

 By ? on Sunday, December 29, 2002 - 09:13 am: Edit

how do you even know that ratio?

 By vp on Sunday, December 29, 2002 - 10:10 am: Edit

using similar triangle properties

 By Pat57575 (Pat57575) on Sunday, December 29, 2002 - 11:59 am: Edit

caltech, math800, good job guys, that's correct.

Did you two set up right triangles using the altitude of the isosceles triangle? That's how I solved it.

An easier way to solve it just in case you were curious...

r= ([s-a][s-b][s-c]/s)^(1/2)
where s is the semiperimeter, a, b, and c are the lengths of the sides

that quantity squared times pi gives you the area.

 By Pat57575 (Pat57575) on Sunday, December 29, 2002 - 03:50 pm: Edit

using the altitude approach I got 10^2= 4^2 + ([6^2 + r^2]^(1/2)+r)^2

solving was somewhat tedious, but, sure enough, I ended up with r = (48/7)^(1/2)

caltech07, you're the first person I've heard from that got into caltech early... congrats!

 By caltech07 on Sunday, December 29, 2002 - 04:25 pm: Edit

Pat, thank you very much.

The approach you give, I spied it on a certain geometry site. I am curious, is it obtained using one of the methods we have been using (altitude, similar triangles, etc.) or is it more clever? Unless one gets lucky formulas like that aren't so easy to memorize for every case, so I just want to sniff of the cleverness through which that was derived.

Thanks!

 By Mitwannabe (Mitwannabe) on Sunday, December 29, 2002 - 05:03 pm: Edit

ic, the reason you can't go ao/r is because the perpendicular line from O to the side of a trangle would make AO the hyp., in which the case of 10/4 would be equal to AO/r, where is the other case you're making the radius the shortest side of the triangle, parallel to the 8 length bottom which gives you a different hyp.

 By 800m on Sunday, December 29, 2002 - 05:19 pm: Edit

Pat, actually I think there is something wrong with your formula. When you plug in a,b,c and s, you don't get r=(48/7)^1/2. [(28-10)(28-10)(28-8)/28]=1620/7 so that can't be right. However, I think I was able to derive the formula. And here it is: (I think it is 1/2*[(s-2a)(s-2b)(s-2c)/s]^1/2)

You have triangle ABC,AB=A, AC=B, BC=CLet x=(a+c-b)/2

Now comes to the last part. r=x*tan<A/2
=[(a+c-b)/2]*[(b^2-(a-c)^2)/((a+c)^2-b^2)]^1/2
={[(b+a-c)(b-a+c)]/[(a+b+c)(a+c-b)]*(a+c-b)^2/4}^1/2
=[(a+b-c)(b+c-a)(a+c-b)/4s]^1/2
a+b-c=s-2c,b+c-a=s-2a,a+c-b=s-sb
so
r=[(s-2a)(s-2b)(s-2c)/4s]^1/2=1/2*[(s-2a)(s-2b)(s-2c)/s]^1/2
And I think that's correct because when you plug in a=b=10 and c=8 you get r=(48/7)^1/2

 By 800m on Sunday, December 29, 2002 - 05:23 pm: Edit

Pat, actually I think there is something wrong with your formula. When you plug in a,b,c and s, you don't get r=(48/7)^1/2. [(28-10)(28-10)(28-8)/28]=1620/7 so that can't be right. However, I think I was able to derive the formula. And here it is: (I think it is 1/2*[(s-2a)(s-2b)(s-2c)/s]^1/2)

You have triangle ABC,AB=A, AC=B, BC=C
Let x=(a+c-b)/2

Now comes to the last part. r=x*tan<A/2
=[(a+c-b)/2]*[(b^2-(a-c)^2)/((a+c)^2-b^2)]^1/2
={[(b+a-c)(b-a+c)]/[(a+b+c)(a+c-b)]*(a+c-b)^2/4}^1/2
=[(a+b-c)(b+c-a)(a+c-b)/4s]^1/2
a+b-c=s-2c,b+c-a=s-2a,a+c-b=s-sb
so
r=[(s-2a)(s-2b)(s-2c)/4s]^1/2=1/2*[(s-2a)(s-2b)(s-2c)/s]^1/2
And I think that's correct because when you plug in a=b=10 and c=8 you get r=(48/7)^1/2

 By 800m on Sunday, December 29, 2002 - 05:24 pm: Edit

Sorry I posted it twice but there is still this thing missing (I don't know why that keeps happening):

Let X,Y,Z be the points where the circle is tangent to the sides of the triangle. X is on BC, Y is on AB, and Z is on AC. BX=x,AY=Y,CZ=z
x+z=c
x+y=a
y+z=b
=>x=(a+c-b)/2

 By Pat57575 (Pat57575) on Sunday, December 29, 2002 - 06:45 pm: Edit

800m, remember the semiperimeter (s) is half of the perimeter... in this case (1/2)(28)=14. Plugging that into the equation yields (48/7)^(1/2)
That's really impressive that you were able to derive the formula using the perimeter though.

caltech07- no, I never derived the formula-- I actually also saw it in a math book. I am amazed that it comes down to such a simple equation (btw, it works with all triangles, not just right and isosceles). If you remember heron's formula, you will notice that the area of the circle is equal to the area of the triangle squared times pi divided by the semiperimeter squared... strange. You might wanna check out 800m's post for a derivation...

 By 800m on Sunday, December 29, 2002 - 06:49 pm: Edit

Oh, sorry I overlooked the word semiperimeter...

 By vp on Monday, December 30, 2002 - 01:15 pm: Edit

HEEEEELLLLLLLLLLLOOOOOOOOOOOOO u guys
is this one big joke or what
who said???????
hehehehheeh
LOL
ROFLOL

 By vp on Monday, December 30, 2002 - 01:19 pm: Edit

ROLLING ON THE FLOOORRRRRRRRRR
hehehehhehee
u can go on and on about ur great semi-perimeter/trig/identity/ and...ahemm...co-ordinate plane or whatever..this is just SIMPLE triangles...similar triangles...the answer is NOT 48/7pi
its
256pi/49

 By BritGirl on Monday, December 30, 2002 - 01:49 pm: Edit

Pat- ur semiperimeter formula is called Hero's formula, derived ages ago. Good job guys, wish I read the question earlier, would have had fun solving it Got any more Pat?

 By vp on Monday, December 30, 2002 - 02:26 pm: Edit

ONE SEC......just one sec.....im highly confused......how can R= hero's formula? the heros formula is for calculating AREA of the TRIANGLE....not r.........

 By Pat57575 (Pat57575) on Monday, December 30, 2002 - 08:12 pm: Edit

vp- the answer is 48pi/7... positively... how did you get 256pi/49?

britgirl- the radius equation is very similar to heron's formula aka hero's formula... but different.

hero's- A=(s[s-a][s-b][s-c])^(1/2)
the other one- r=([s-a][s-b][s-c]/s)^(1/2)

as you may have noticed, I commented on the similarities above.

P.S. yes, I have some other good ones-- I'll post another soon.

 By dfawcett on Monday, February 17, 2003 - 05:22 pm: Edit

The answer is 28/3 pi. Area of triangle = pr, where p is semiperimeter and r is radius of inscribed circle. Using pythagorean thm, the height of the triangle is sqroot of 84. Therefore, .5bh = pr gives an r = 2/3 x sqroot 21. So area of circle = 28/3 pi