Question

Upon the request of @thomas5267 I've chosen a problem.

Answer 1

May not be a difficult one. Just try~ :)

Answer 2

Note: I just picked it randomly :(

Answer 3

lol you avoided the verbs and adjectives =)))) it was just a joke :P

Answer 4

I'm not sure if it is a cute one and if others can enjoy it :P

Answer 5

BTW, no more problems after tomorrow until the end of April, the beginning of May :)

Answer 6

G is the circumcentre of ABC so segment BS is a median?

Answer 7

I was thinking if BS is perpendicular to AC
circumcentre is constructed by perpendicular bisectors

Answer 8

@Callisto then its connected to vertex B then it is an isosceles triangle?

Answer 9

@cuty_shai2000 which triangle?

Answer 10

ABC with B as the vertex angle

Answer 11

AO and SC are both perpendicular to BC, so they are paralllel
CG and SA are both perpendicular to AB, so they are paralllel
SO AHCS is a paralleogram
G is the midpoint of BS, so R is the midpoint of BC and GR = 1/2 SC

Answer 12

@cuty_shai2000 I'm not sure if we could draw such conclusion

Answer 13

For 16b(ii).
\[
\begin{align*}
AB&=\sqrt{180} \\
AC&=\sqrt{160} \\
BC&=10 \\
\end{align*}
\]
From the formula from Wikipedia, http://en.wikipedia.org/wiki/Circumscribed_circle.
\[
\begin{align*}
\text{diameter}&=\frac{2abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}} \\
d&=\frac{2(\sqrt{180})(\sqrt{160})(10)}{\sqrt{(\sqrt{180}+\sqrt{160}+10)(-\sqrt{180}+\sqrt{160}+10)(\sqrt{180}-\sqrt{160}+10)(\sqrt{180}+\sqrt{160}-10}} \\
d&=14.14214
\end{align*}
\]
I have enough of this, lunch now, cya.

Answer 14

Please finish my calculation of 16b(ii).

Answer 15

@eliassaab i agree with you ..

Answer 16

CH i think .. not CG

Answer 17

The equation of the circle is
\[(x+1)^2+(y-5)^2=50
\]

Answer 18

im confused with G as the circumcentre

Answer 19

@eliassaab how did you come up with that?

Answer 20

H=(0,2)

Answer 21

You write the eqaution of the circle as
\[(x-a)^2+(y-b)^2=d
\]
You find three eqaution with 3 unknowns by saying that the circle goes through the three points
You solve for a, b and d

Answer 22

The three equations are
\[
a^2+b^2-24 b-d+144=0\\
a^2+12 a+b^2-d+36=0\\
a^2-8 a+b^2-d+16=0
\]

Answer 23

First Equation - Second, you get
\[-12 a-24 b+108=0
\]
First Equation - Third, you get
\[
8 a-24 b+128=0
\]
Solve you get a= -1 and b=5, substitute a and b in one of the original equation to get
d=50

Answer 24

Why \((x-a)^2+(y-b)^2)=d\)? The standard form of circle should be \((x-h)^2+(y-k)^2=r^2\) where (h, k) is the center of the circle and r is the radius. Again, I haven't learn this topic so I am pretty unsure about that.

Answer 25

So
\[r=\sqrt {50}
\]

Answer 26

How do you find the coordinates of h then?

Answer 27

that's the question i want to ask earlier ..

Answer 28

H is (0, y) as AO is a straight line. That's all I know.

Answer 29

If B, O , H and G are concyclic, then BH must be the diameter since angle BOH=90.
So the center is in the middle of BH and so its coordinates are (-3,1).
The radius is 1/2 BH
\[
BH^2 = 36 + 4 = 40\\
BH = 2 \sqrt {10}\\
r = \sqrt {10} \\
(x+3)^2 + (y-1)^2 = 10
\]
G=(-1,5) does not belong to that circle
\[(x+3)^2 + (y-1)^2 = 10

\]
So B,O, H and G are not concyclic

Answer 30

@thomas5267
I put h=a and k= b in my eqautions and solved for a, b and d.
See the details above,

Answer 31

Some of you asked why H=(0,2).
Here is a proof:

OA =12
AH=SC= 2 GR= 2 (5)=10
OH= OA- AH= 12 -10=2