Three Points Casey's Theorem

---

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

---

Solution

References

1. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 2, Moscow, 1952, #123(a).

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

The problem is just a particular instance of Casey's theorem, wherein three of the four circles degenerate into points. Still, I'll give a proof independent of Casey's theorem.

Point M serves the center of similitude of the two circles. If A', B', C' are the second points of intersection of AM, BM, CM, with (S) then

A'M/AM = B'M/BM = C'M/CM = ρ/r

where r and ρ are the radii of the circumcircle (ABC) and that of (S), respectively. It follows that

AA'/AM = BB'/BM = CC'/CM = (r ± ρ)/ρ.

(The sign is chosen depending on whether the circumcircle (ABC) is outside or inside (S).)

On the other hand, by the Power of a Point Theorem,

AM·AA' = (AT_a)²,
BM·BB' = (BT_b)²,
CM·CC' = (CT_c)²,

such that

AM² / (AT_a)² = AM / AA',
BM² / (BT_b)² = BM / BB',
CM² / (CT_c)² = CM / CC',

implying

AM / AT_a = BM / BT_b = CM / CT_c = √ρ / (r ± ρ),

which simply shows that the required identity

BC·AT_a + AC·BT_b = AB·CT_c.

is a direct consequence of that of Ptolemy's applied to the quadrilateral ABCM:

BC·AM + AC·BM = AB·CM.

Ptolemy's Theorem

1. Ptolemy's Theorem
2. Sine, Cosine, and Ptolemy's Theorem
3. Useful Identities Among Complex Numbers
4. Ptolemy on Hinges
5. Thébault's Problem III
6. Van Schooten's and Pompeiu's Theorems
7. Ptolemy by Inversion
8. Brahmagupta-Mahavira Identities
9. Casey's Theorem
10. Three Points Casey's Theorem
11. Ptolemy via Cross-Ratio

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|