1. Formulas Notations: The area of ABC will be denoted by [ABC] or SABC.

As usual, we let a, b, c be the lengths of sides BC, CA, AB respectively.

B a O A D C

Extended Sine Law. Let R be the radius of the circumcircle of ABC. Then a b c = = = 2 R. sin A sin B sin C
a a = = BD = 2R. sin A sin D

Proof. Draw diameter BD. Then

A c b

ha a A

Area of Triangle. Letting ha denote the height from A to side BC, we have
[ ABC ] =
C

ah a ab sin C abc = = . 2 2 4R

c r

I r r

Let s = (a+b+c)/2 be the semiperimeter of ABC. Let I be the incenter of ABC and r be the radius of the incircle of ABC. Then
C

[ ABC ] = [ AIB ] + [ BIC ] + [CIA] =

ar br cr + + = sr. 2 2 2

Herons Formula. [ ABC ] = s ( s a)(s b)(s c) .

Proof. By cosine law, cos C =(a2+b2c2)/(2ab). We have
a 2b 2 a 2b 2 2 [ ABC ] = sin C = (1 + cos C )(1 cos C ) 4 4 (2ab + a 2 + b 2 c 2 )(2ab a 2 b 2 + c 2 ) = 16 ( a + b + c ) ( a + b c ) (c + a b ) (c a + b ) = 2 2 2 2 = s ( s c)(s b)(s a).
2

(Remarks. Combining with the formulas above, we have abc ( s a)( s b)( s c) R= r= and .) s 4 s ( s a )(s b)( s c)
A M N

Example 1. (2005 APMO) In a triangle ABC, points M and N are on sides AB and AC, respectively, such that MB=BC=CN. Let R and r denote the circumradius and the inradius of the triangle ABC, respectively. Express the ratio MN/BC in terms of R and r.

Solution. (Due to Tsoi Yun Pui) Let AB=x+y, BC=y+z, CA=z+x (that is,
x= AB + CA BC = s a, 2
y=

CA + BC AB BC + AB CA = s c ). = s b and z = 2 2

Then AM=xz and AN=xy. By cosine law, MN2=AN2+AM22ANAM cos A and cos A=(AB2+AC2BC2)/2ABAC. Expressing AN, AM, AB, AC, BC in terms of x, y, z, we get
2 x( z y )( z 2 y 2 ) + ( x z )( x y )( y + z ) 2 MN = . ( x + y )( z + x)
2

MN 2 2 x( z y )( z 2 y 2 ) + ( x z )( x y )( y + z ) 2 8 xyz = = 1 . Then 2 2 ( x + y )( y + z )( z + x) BC ( x + y )( z + x)( y + z )

Now r2 =

( s a )( s b)( s c ) xyz = s s

and sr =[ABC] =

abc ( x + y )( y + z )( z + x ) = . Then 4R 4R

MN 8r 2 s 2r = 1 = 1 . BC R 4 Rsr

A c m B D
m+n = a

b p n C

Stewarts Theorem. Let D be a point inside BAC (<180). Let p, m and n be the lengths of line segments AD, BD and CD respectively. Then
D is on segment BC b2m +c2n = a(p2+mn).

Proof.

D is on segment BC ADB +ADC=180 cosADB + cosADC = 0

n2 + p2 b2 m2 + p 2 c2 =cosADB = cosADC = 2mp 2np

b2m +c2n = a(p2+mn). Formulas. (1) If AD is the median to side BC, then m = a/2 = n. Let ma denote 1 2b 2 + 2c 2 a 2 . This formula the length of AD. Stewarts theorem yields ma = 2 is sometimes refered to as Apollonius formula. Note we have the interesting formula 2 2 4( m a + mb + m c2 ) = 3( a 2 + b 2 + c 2 ).
(2) If AD is the angle bisector of BAC, then m/n=c/b and m+n=a imply m=ca/(b+c) and n=ba/(b+c). Let ta denote the length of AD. Stewarts theorem
a2 yields t a = bc1 (b + c) 2 .

Subtended Angle Theorem. Let D be a point inside BAC (<180). Let =BAD and =CAD. Then
C

D is on segment BC

sin( + ) sin sin = + . AD AC AB

Proof.

D is on segment BC [ABC]=[BAD]+[CAD] AB AC sin( + ) AB AD sin AC AD sin = +

sin( + ) sin sin = + . AD AC AB

D B G G' C F E

Example 2. (1999 Chinese National Math Competition) In convex quadrilateral ABCD, diagonal AC bisects BAD. Let E be on side CD such that BEAC=F and
DFBC=G. Prove that GAC=EAC.

Solution. Let BAC=DAC= and G be on segment BC such that GAC = EAC=. We will show G, F, D are collinear, which implies G=G. Applying the subtended angle theorem to ABE, ABC and ACD respectively, we get
(1)
sin( + ) sin sin = + AF AB AE

, (2)

sin sin sin( ) = + AG ' AB AC

and (3)

sin sin sin( ) = + . AE AD AC

Doing (1)(2)+(3), we get

C A N

sin( + ) sin sin = + . Then G, F, D are collinear. AF AD AG '

P
Q R O

Example 3. (Butterfly Theorem) Let A,C,E,B,D,F be

B

points in cyclic order on a circle and CDEF=P is the midpoint of AB. Let M = ABDE and N = ABCF. Prove that MP = NP.

Solution. By the intersecting chord theorem, PCPD=PEPF, call this x. Applying the subtended angle theorem to PDE and PCF, we get
sin( + ) sin sin sin( + ) sin sin = + = + and . Subtracting these equations, PM PE PD PN PF PC 1 PF PE PD PC 1 sin . Let Q, R be midpoints = sin we get sin( + ) x x PM PN

of EF, CD respectively. Since OPAB, we have

and
Then sin( + )

PF PE = 2PQ = 2OP cos(90o ) = 2OP sin PD PC = 2 PR = 2OP cos(90 o ) = 2OP sin .

1 1 o = 0. Since 0 < + < 180 , we get PM =PN. PM PN

Exercises
1. Let a, b, c denote the lengths of the sides BC, CA, AB respectively. Let ha, hb, hc be the heights from A, B, C to the opposite sides respectively. Let R be the circumradius, r be the inradius and s be the semiperimeter of ABC. 1 1 1 1 (a) For ABC, show that h + h + h = r . a b c (b) Show that r = 4 R sin
A r A B C .) sin sin . (Hint: Show tan = 2 2 2 2 sa

2. Show that among all triangles with the same perimeter, the equilateral triangle has the largest area. 3. (1996 Iranian Math Olympiad) Let ABC be a scalene triangle (i.e. no two sides equal). The medians from A, B, C meet the circumcircle again at L, M, N respectively. If LM=LN, prove that 2BC2=AB2+AC2. (Hint: Show LN LG = first.) AC CG 4. Let a,b,c and a, b, c be the lengths of two triangles. Let K(x,y,z) be the area of a triangle with side lengths x,y,z. Show that
K (a + a' , b + b' , c + c' ) K (a, b, c) + K (a' , b' c' ) . When does equality hold?

5. Let G be the centroid of ABC. A line through G intersects sides AB, AC at E, F respectively. Prove that EG 2GF. (Hint: Draw medians BM and CN.) 6. AB is a diameter of a circle and C is a point on the circle. The tangents to the circle at A and at C intersect at P. Let D be the foot of perpendicular from C to AB. Let Q be the midpoint of CD. Prove that P, Q, B are collinear. (Hint: Consider CPB.)

2. Power of Points Respect to Circles

The power of a point P with respect to a circle is the number d2 r2, where d is the distance from P to the center of the circle and r is the radius of the circle. (If P is outside the circle, the power is positive. If P is inside, the power is negative. If P is on the circle, the power is 0.)
P O A r-d r+d

A'

In the intersecting chord theorem, if P is inside a circle and AA' is a chord through P, then the product PA PA' is constant and can be determined by taking the case the chord AA' passes through P and the center O. This gives PA PA' = r2 d2, where r is the radius of the circle and d = OP. In the case P is outside the circle, the product PA PA' can be determined by taking the limiting case when PA is tangent to the circle. Then PA PA' = d2 r2.

A=A' d P

r O

Thus, PA PA' is the absolute value of the power of P with respect to the circle. This is known as the power-of-a-point theorem. Next we will look at points having equal power with respect to two circles. Theorem. On a plane, for distinct points R,S and a number m, the locus of all points X such that RX2SX2 = m is a line perpendicular to line RS. Also, for distinct points R,S,X,Y, we have RX2SX2 = RY2SY2 if and only if RS XY.
X=(x,y) R=(0,0) S=(b,0)

Proof. Let R, S have coordinates (0,0), (b,0) respectively. Point X with coordinates (x,y) is on the locus if and only if (x2+y2) ((bx)2+y2) = m, which yields the line x = (m+b2)/2b perpendicular to RS.

The second statement follows by using Pythagoras theorem for the if-part and letting m = RY2SY2 so that X,Y are both on the locus for the only-if-part. Let circles C1 and C2 have distinct centers O1 and O2. By the theorem, the points X whose powers with respect to C1 and C2 are equal (i.e. O1X2 r12 = O2X2 r22 ) form a line perpendicular to line O1O2. This line is called the radical axis of the two circles.
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For three circles C1, C2, C3 with distinct centers, if their centers O1, O2, O3 are collinear, then the three radical axes of the three pairs of circles are parallel or coincide. Otherwise, the radical axes intersect at a point called the radical center of the three circles. (This is because the intersection point of any two of these radical axes has equal power with respect to all three circles, hence it is on the third radical axis too.) If two circles C1 and C2 intersect, their radical axis is the line through the intersection point(s), which is C2 perpendicular to the line of the centers. (This is O1 because the intersection point(s) have 0 power with O2 C1 respect to both circles, hence they are on the radical axis.)
P C1 O1 C3 O2 C2

If the two circles do not intersect, their radical axis can be found by taking a third circle C3 intersecting both C1 and C2. Let the radical axis of C1 and C3 intersect the radical axis of C2 and C3 at P. Then the radical axis of C1 and C2 is the line through P perpendicular to the line of centers of C1 and C2.

A F B C D
C3 F A E B C D C1 C2

Example 1. (1997 USA Math Olympiad) Let ABC be a triangle, and draw isosceles triangles BCD, CAE, ABF externally to ABC, with BC, CA, AB as their respective bases. Prove the lines through A, B, C, perpendicular to the lines EF, FD, DE, respectively, are concurrent. Solution 1. Let C1 be the circle with center D and radius BD, C2 be the circle with center E and radius CE, and C3 be the circle with center F and radius AF. The line through A perpendicular to EF is the radical axis of C2 and C3, the line through B perpendicular to FD is the radical axis of C3 and C1 and the line through C perpendicular to DE is the radical axis of C1 and C2. These three lines concur at the radical center of the three circles.
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P B D

Solution 2. Let P be the intersection of the perpendicular line from B to FD with the perpendicular E line from C to DE. Then PBFD and PCDE. By the theorem above, we have
C

PF2PD2= BF2BD2 and PD2PE2= CD2CE2.

Adding these and using AF=BF, BD=CD and CE=AE, we get PF PE = AF AE2. So PAEF and P is the desired concurrent point.
2 2 2

Example 2. (Butterfly Theorem) Let A,C,E,B,D,F be points in cyclic order on a

circle and CDEF=P is the midpoint of AB. Let M = ABDE and N = ABCF. Prove that MP = NP.
G

Solution. Let G=CFDE. Applying Menelaus theorem to lines CD and EF cutting GMN, we have
E

C A N P

NP MD GC NP ME GF = 1. = 1 and PM EG FN PM DG CN
B

Multiply these equations and using GFGC=EGDG from the intersecting chord theorem, we get
D

NP 2 FN CN AN NB ( AP NP)( AP + NP) AP 2 NP 2 = = = = . PM 2 ME MD BM AM ( BP MP)( BP + MP) BP 2 MP 2

Since AP=BP, we get MP=NP by cross-multiplying the leftmost and rightmost parts of the equation.

Remarks. Viewing AB as a diameter of the circle centered at P and radius AP, we can get ANNB=AP2 NP2 by the power-of-a-point theorem.
B K N O A C P M

Example 3. (1985 IMO) A circle with center O passes through vertices A and C of a nonisosceles triangle ABC and intersects side AB at K and side BC at N. Let the circumcircles of triangles ABC and KBN intersect at B and M. Prove that OM is perpendicular to BM.

Solution. Since ABC is not isosceles, so lines AC and KN are not parallel. For the three circles mentioned, the radical axes of the three pairs are lines AC, KN and BM. (Their centers are noncollinear because two of them are on the perpendicular bisector of AC and two of them are also on the perpendicular bisector of KN.) So the axes will concur at the radical center P. Since

PMN = BKN = NCA,

it follows that P, M, N, C are concyclic. By the power of a point theorem, and BM BP = BN BC = BO2 r2 PM PB = PN PK = PO2 r2, PO2 BO2 = BP(PM BM) = PM2 BM2. By the theorem above, this implies OM PB, which is the same as OM BM.

where r is the radius of the circle through A, C, N, K. Then

Example 4. (1997 Chinese Math Olympiad) Let quadrilateral ABCD be inscribed in a circle. Suppose lines AB and DC intersect at P and lines AD and BC intersect at Q. From Q, construct the tangents QE and QF to the circle, where E and F are the points of tangency. Prove that P, E, F are collinear.
A E D Q M P O1 C F B C1

Solution. Let the circumcircle of QCD intersect PQ at M. Since PMC= QDC= ABC, points B, C, M, P are concyclic. Let r1 be the radius of the circumcircle C1 of ABCD and O1be the center of C1. By power of a point,
and PO12 r12 = PC PD = PM PQ QO12 r12 = QC QB = QM PQ.

Then

PO12 QO12 = (PM QM) PQ = PM2 QM2

implies PQO1M. Since QMO1=QEO1=QFO1=90, the circle C2 with QO1 as diameter passes through M, E, F. Since C1, C2 intersect at E, F, line EF is their radical axis. If r2 is the radius of C2 and O2 is the center of C2, then PO12 r12 = PM PQ = PO22 r22. So P lies on the radical axis of C1, C2, which is the line EF.

F B

O H D N E C

Example 5. (2001 Chinese National Senior High Math Competition) As in the figure, in ABC, O is the circumcenter. The three altitudes AD, BE and CF intersect at H. Lines ED and AB intersect at M. Lines FD and AC intersect at N. Prove that (1) OBDF and OCDE; (2) OHMN.

Solution 1. (1) Since AFC = 90=ADC, so A,C,D,F are concyclic. Then BDF =BAC. Also, OBC =(180BOC) = 90BAC = 90BDF
implies OBDF. Similarly, OCDE. (2) We have CHMA MC2MH2 = AC2AH2 BHNA NB2NH2 = AB2AH2 DABC DB2DC2 = AB2AC2 OBDF = DN BN2BD2 = ON2OD2 OCDE =DM CM2CD2 = OM2OD2. (a) (b) (c) (d) (e)

Doing (a)(b)+(c)+(d)(e), we get NH2MH2 = ON2OM2. So OHMN.

Solution 2. (1) Let J be on line DM such that JBDF. Since AFC = 90= ADC, so A,C,D,F are concyclic. Then JBD = BDF = BAC. So line BJ is tangent to the circumcircle of ABC. Then OBJB. So OBDF. Similarly, OCDE.
(2) Recall the Euler line OH of ABC contains the center of the circumcircle and the center of the nine point circle. We will show N, M are on the radical axis of these circles and hence OHMN. From (1), we know A,C,D,F are concyclic. By the intersecting chord theorem, NANC = NFND. Since AC is a chord of the circumcircle and FD is a chord on the nine-point circle, so N is on the radical axis of the circles and similarly for M.

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Exercises
1. Show that OI2=R22Rr, where O is the circumcenter, R is the circumradius and r is the inradius of ABC. 2. (1995 Russian Math Olympiad) ABCD is a quadrilateral such that AB=AD and ABC and CDA are right angles. Points F and E are chosen on BC and CD respectively so that DFAE. Prove that AF BE. (Hint: Show AB2AE2 =FB2 FE2.) 3. (2005 CMO Summer Camp A-Level Test) In acute ABC, AB=AC and P is a point on ray BC. Points X and Y are on rays BA and AC such that PX||AC and PY||AB. Point T is the midpoint of minor arc BC on the circumcircle of ABC. Prove that PTXY. (Hint: Show PX2PY2 =TX2 TY2.) 4. A circle with center O is inscribed in quadrilateral ABCD. Let K, L, M, N be the tangent points on sides AB, BC, CD, DA respectively. Let S = KLMN. Prove that BDSO. (Hint: Show BS2BO2 =DS2 DO2.) 5. (2005 CMO Team Selection Test) Let ABCD be a cyclic quadrilateral. Let P=ACBD. A circle C1 passing through P and B intersects another circle C2 passing through P and A at a point Q (other than P). Let E and F be the intersections of the circumcircle of ABCD with C1 and C2 respectively. Prove that lines PQ, CE and DF are concurrent or they are parallel. (Hint: Let I be the intersection of line CE with C1 and J be the intersection of line DF with C2. Show that I, P, J are collinear and E,F,J,I are concyclic.) 6. (2000 Russian Math Olympiad) Let E be on the median CD of triangle ABC. Let S1 be the circle passing through E and tangent to line AB at A, intersecting side AC again at M; let S2 be the circle passing through E and tangent to line AB at B, intersecting side BC again at N. Prove that the circumcircle of triangle CMN is tangent to circles S1 and S2. (Hint: Show DE is the radical axis of S1 and S2. Show A,B,M,N are concyclic.)

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3. Miscellaneous Examples Example 1. (1994 Canadian Math Olympiad) Let ABC be an acute triangle. Let D be on side BC such that ADBC. Let H be a point on the segment AD different from A and D. Let line BH intersect side AC at E and line CH intersect side AB at F. Prove that EDA =FDA.
P F A E C Q

Solution 1. Draw a line through A parallel to BC. Let the line intersect line DE at Q and line DF at P. Note APF~BFD. So AP/BD=AF/BF. Similarly,
AQ/CD=AE/CE. Since H=ADBECF, applying Cevas theorem to ABC, we get
AF AE AF BD CE BD = CD AP=AQ. = 1 FB CE FB DC EA

This with DA=DA and DAP = 90=DAQ yield DAPDAQ. Therefore, EDA =FDA. A Solution 2. Let A, E be the mirror image of A, E
F H E C E' A'

with respect to line BC. Since AFCDEH = B, applying Cevas theorem to AHC, we have
AD HF CE A' D HF CE ' = . DH FC EA DH FC E ' A' By the converse of Menelaus theorem, D, F, E are collinear. Hence, EDA =EDA=FDA.

1=

Example 2. (1997 USA Math Olympiad) Let ABC be a triangle, and draw isosceles triangles BCD, CAE, ABF externally to ABC, with BC, CA, AB as their respective bases. Prove the lines through A, B, C, perpendicular to the lines EF, FD, DE, respectively, are concurrent.
A F B A' F' E' B' D' C' D

Solution. Let A, B, C be points on FE, DF, ED respectively such that AAFE, BBDF and CC E ED. Let D,E,F be points on CB,AC,BA respectively such that DD CB, EE AC and FF BA. Now DD, EE, FF are perpendicular bisectors of the C sides of ABC. So they concur. Applying the trigonometric form of Cevas theorem to DEF,
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sin E ' EF sin F ' FD sin D ' DE = 1. sin DEE ' sin EFF ' sin FDD ' Since EE CA and EF AA, so EEF CAA. Similarly, DEE

AAB, FFDABB, EFFBBC, DDEBCC and FDD BBC. So sin CAA' sin ABB ' sin BCC ' = 1. sin A' AB sin B ' BC sin B ' BC By the converse of Cevas theorem, we get AA, BB, CC are concurrent, which is the required conclusion.

Example 3. (1996 IMO) Let ABCDEF be a convex hexagon such that AB is parallel to DE, BC is parallel to EF and CD is parallel to FA. Let RA, RC, RE denote the circumradii of triangles FAB, BCD, DEF, respectively and let P denote the perimeter of the hexagon. Prove that RA + RC + RE P/2.
P A D S F E R B C Q

Solution. Due to the parallel opposite sides, we have A =D, B =E, C =F. Let PQRS be the smallest rectangle containing the hexagon with side BC on PQ as shown. We have BF PS=QR. So
2BF PS+QR = AP+AS+DQ+DR = (AB sin B+FA sin C)+(CD sin C+DE sin B).

Similarly, and

2DB (CD sin A+BC sin B)+(EF sin B+FA sin A) 2FD (EF sin C+DE sin A)+(AB sin A+BC sin C).

BF DB FD , RC = RE = . Dividing and 2 sin A 2 sin C 2 sin B the first inequality by 4sin A, second inequality by 4sin C, third inequality by 4sin E and adding them, we get by the AM-GM inequality that

By the extended sine law, R A =

RA + RC + RE

The result follows.

sin B sin A 1 sin C sin B 1 + + + BC + AB 4 sin A sin B 4 sin B sin C 1 (AB+BC+CD+DE+EF+FA) = P/2. 2

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