- 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,

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,

Vijay Prasad Nalluri.

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

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[Non-text portions of this message have been removed] - Dear Vijaya Prasad Nalluri,

In Hyacinthos message #8891, you wrote:

>> Dear Darij Grinberg,

Yes, thanks. This was nearly the same way I proved

>> 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

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