## Tangents to the circumcircle

Expand Messages
• A little warm-up problem for 2004: Let a line l touch the circumcircle of a triangle ABC at a point on the arc BC. Denote by x, y, z the distances from A, B, C
Message 1 of 3 , Dec 31, 2003
A little warm-up problem for 2004:

Let a line l touch the circumcircle of a triangle
ABC at a point on the arc BC. Denote by x, y, z
the distances from A, B, C to l. Prove, without the
use of coordinate systems, that

a * sqrt x = b * sqrt y + c * sqrt z.

Happy New Year to all!

Darij Grinberg
• Dear Darij Grinberg, Let line l touch circle ABC at P.Let Q be the antipode of P.From similar triangles PAX,QPA we have x/AP = AP/2R giving AP^2 = 2xR,
Message 2 of 3 , Dec 31, 2003
Dear Darij Grinberg,
Let line l touch circle ABC at P.Let Q be the antipode of P.From similar triangles PAX,QPA we have
x/AP = AP/2R giving AP^2 = 2xR,
similarly, we get BP^2 = 2yR and CP^2 = 2zR
So BP/AP = sqrt (y/x) and CP/AP = sqrt (z/x)
From Ptolemy's theorem applied to the cyclic quadrilateral ABPC,
BC * AP = AC* BP + AB * CP
Dividing either side by AP
BC = AC * (BP/AP) + AB * (CP/AP)
a = b * sqrt (y/x) + c * sqrt (z/x)
a * sqrt x = b *sqrt y + c * sqrt z
With warm NY Gtngs to all,

Darij Grinberg <darij_grinberg@...> wrote:
A little warm-up problem for 2004:

Let a line l touch the circumcircle of a triangle
ABC at a point on the arc BC. Denote by x, y, z
the distances from A, B, C to l. Prove, without the
use of coordinate systems, that

a * sqrt x = b * sqrt y + c * sqrt z.

Happy New Year to all!

Darij Grinberg

---------------------------------

To visit your group on the web, go to:
http://groups.yahoo.com/group/Hyacinthos/

To unsubscribe from this group, send an email to:
Hyacinthos-unsubscribe@yahoogroups.com

---------------------------------
Do you Yahoo!?
Find out what made the Top Yahoo! Searches of 2003

[Non-text portions of this message have been removed]
• Dear Vijaya Prasad Nalluri, ... Yes, thanks. This was nearly the same way I proved my formula. It is related to trilinear coordinates, namely: If a line has
Message 3 of 3 , Dec 31, 2003

In Hyacinthos message #8891, you wrote:

>> Dear Darij Grinberg,
>> Let line l touch circle ABC at P.Let Q be the
>> antipode of P.From similar triangles PAX,QPA we
>> have x/AP = AP/2R giving AP^2 = 2xR,
>> similarly, we get BP^2 = 2yR and CP^2 = 2zR
>> So BP/AP = sqrt (y/x) and CP/AP = sqrt (z/x)
>> From Ptolemy's theorem applied to the cyclic
>> BC * AP = AC* BP + AB * CP
>> Dividing either side by AP
>> BC = AC * (BP/AP) + AB * (CP/AP)
>> a = b * sqrt (y/x) + c * sqrt (z/x)
>> a * sqrt x = b *sqrt y + c * sqrt z

Yes, thanks. This was nearly the same way I proved
my formula. It is related to trilinear coordinates,
namely: If a line has trilinear equation
kx + ly + mz = 0, then it is tangent to the
circumcircle of triangle ABC if and only if

(ak)^2 + (bl)^2 + (cm)^2
= 2 ( blcm + cmak + akbl ),

or, equivalently,

( sqrt(bl) + sqrt(cm) - sqrt(ak) )
* ( sqrt(cm) + sqrt(ak) - sqrt(bl) )
* ( sqrt(ak) + sqrt(bl) - sqrt(cm) )
* ( sqrt(ak) + sqrt(bl) + sqrt(cm) ) = 0.

Now, since the last factor never vanishes, we get
the condition that one of the values sqrt(ak),
sqrt(bl), sqrt(cm) equals the sum of the two
others.

Now, if x, y, z are the distances from A, B, C to
the line, the line coordinates k, l, m are
proportional to ax, by, cz, hence we get our
formula above.

Sincerely,
Darij Grinberg
Your message has been successfully submitted and would be delivered to recipients shortly.