Download Angle Bisectors Bisect Arcs and more Summaries Mathematics in PDF only on Docsity!
Volume 11, Number 2 April 2006 – May 2006
Angle Bisectors Bisect Arcs
Kin Y. Li
Olympiad Corner
Below was the Find Round of the 36th Austrian Math Olympiad 2005.
Part 1 (May 30, 2005) Problem 1. Show that an infinite number of multiples of 2005 exist, in which each of the 10 digits 0,1,2,…, occurs the same number of times, not counting leading zeros. Problem 2. For how many integer values of a with | a | ≤ 2005 does the system of equations x^2 = y + a , y^2 = x + a have integer solutions? Problem 3. We are given real numbers a , b and c and define sn as the sum sn = an
- bn^ + cn^ of their n -th powers for non-negative integers n. It is known that s 1 = 2, s 2 = 6 and s 3 = 14 hold. Show that
| s^2 n^ − sn − 1 ⋅ sn + 1 |= 8
holds for all integers n > 1. Problem 4. We are given two equilateral triangles ABC and PQR with parallel sides, “one pointing up” and “one pointing down.” The common area of the triangles’ interior is a hexagon. Show that the lines joining opposite corners of this hexagon are concurrent.
(continued on page 4)
Editors: Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK ଽ υ ࣻ (KO Tsz-Mei) గ ႀ ᄸ (LEUNG Tat-Wing) ፱ (LI Kin-Yin), Dept. of Math., HKUST ֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance. On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is August 16, 2006. For individual subscription for the next five issues for the 05-06 academic year, send us five stamped self-addressed envelopes. Send all correspondence to: Dr. Kin-Yin LI Department of Mathematics The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: [email protected]
In general, angle bisectors of a triangle do not bisect the sides opposite the angles. However, angle bisectors always bisect the arcs opposite the angles on the circumcircle of the triangle! In math competitions, this fact is very useful for problems concerning angle bisectors or incenters of a triangle involving the circumcircle. Recall that the incenter of a triangle is the point where the three angle bisectors concur.
Theorem. Suppose the angle bisector of ∠ BAC intersect the circumcircle of ∆ ABC at X ≠ A. Let I be a point on the line segment AX. Then I is the incenter of ∆ ABC if and only if XI = XB = XC. A
B C
X
I
Proof. Note ∠ BAX = ∠ CAX = ∠C BX.
So XB = XC. Then
I is the incenter of ∆ ABC
⇔ ∠ CBI = ∠ABI
⇔ ∠ IBX − ∠C BX = ∠BIX − ∠BAX
⇔ ∠IBX = ∠ BIX
⇔ XI = XB = XC.
Example 1. ( 1982 Australian Math Olympiad ) Let ABC be a triangle, and let the internal bisector of the angle A meet the circumcircle again at P. Define Q and R similarly. Prove that AP
A
B C
P
R
I Q
Solution. Let I be the incenter of ∆ ABC. By the theorem, we have 2 IR = AR + BR
AB and similarly 2 IP > BC , 2 IQ > CA. Also AI + BI > AB , BI + CI > BC and CI + AI > CA. Adding all these inequalities together, we get 2( AP + BQ + CR ) > 2( AB + BC + CA ).
Example 2. ( 1978 IMO ) In ABC , AB = AC. A circle is tangent internally to the circumcircle of ABC and also to the sides AB , AC at P , Q , respectively. Prove that the midpoint of segment PQ is the center of the incircle of ∆ ABC.
A
B C
X
P Q
I
Solution. Let I be the midpoint of line segment PQ and X be the intersection of the angle bisector of ∠ BAC with the arc BC not containing A.
By symmetry, AX is a diameter of the circumcircle of ∆ ABC and X is the midpoint of the arc PXQ on the inside circle, which implies PX bisects
∠ QPB. Now ∠ ABX = 90 ˚ = ∠ PIX
so that X, I, P, B are concyclic. Then
∠ IBX = ∠ IPX = ∠B PX = ∠ BIX.
So XI = XB. By the theorem, I is the incenter of ∆ ABC.
Example 3. ( 2002 IMO ) Let BC be a diameter of the circle Γ with center O. Let A be a point on Γ such that 0˚ < ∠ AOB < 120˚. Let D be the midpoint of the arc AB not containing C. The line through O parallel to DA meets the line AC at J. The perpendicular bisector of OA meets Γ at E and at F. Prove that J is the incenter of the triangle CEF.
C O B
A
E
F
D
J
Solution. The condition ∠ AOB <
120 ˚ ensures I is inside ∆ CEF (when
∠ AOB increases to 120 ˚, I will
coincide with C ). Now radius OA and chord EF are perpendicular and bisect each other. So EOFA is a rhombus. Hence A is the midpoint of arc EAF.
Then CA bisects ∠ ECF. Since OA =
OC , ∠ AOD = 1/2 ∠ AOB = ∠ OAC.
Then DO is parallel to AJ. Hence ODAJ is a parallelogram. Then AJ = DO = EO = AE. By the theorem, J is the incenter of ∆ CEF.
Example 4. ( 1996 IMO ) Let P be a point inside triangle ABC such that
∠ APB − ∠ ACB = ∠ APC − ∠ ABC.
Let D , E be the incenters of triangles APB , APC respectively. Show that AP , BD and CE meet at a point.
A
B C
F
G
P
H
J I
D E
K
Solution. Let lines AP , BP , CP intersect the circumcircle of ∆ ABC again at F , G , H respectively. Now
∠ APB − ∠ ACB = ∠ FPG − ∠ AGB
= ∠ FAG.
Similarly, ∠ APC − ∠ ABC = ∠ FAH.
So AF bisects ∠ HAG. Let K be the
incenter of ∆ HAG. Then K is on AF and lines HK , GK pass through the midpoints I , J of minor arcs AG , AH respectively. Note lines BD , CE also
pass through I , J as they bisect ∠ ABP ,
∠ ACP respectively.
Applying Pascal’s theorem (see vol .10, no. 3 of Math Excalibur ) to B , G , J , C ,
H , I on the circumcircle, we see that P=BG ∩ CH , K=GJ ∩ HI and BI ∩ CJ= BD ∩ CE are collinear. Hence, BD ∩ CE is on line PK , which is the same as line AP.
Example 5. ( 2006 APMO ) Let A , B be two distinct points on a given circle O and let P be the midpoint of line segment AB. Let O 1 be the circle tangent to the line AB at P and tangent to the circle O. Let ℓ be the tangent line, different from the line AB , to O 1 passing through A. Let C be the intersection point, different from A , of ℓ and O. Let Q be the midpoint of the line segment BC and O 2 be the circle tangent to the line BC at Q and tangent to the line segment AC. Prove that the circle O 2 is tangent to the circle O.
A (^) P B
N
L
C Z
Q
M
J
K
Solution. Let the perpendicular to AB through P intersect circle O at N and M with N and C on the same side of line AB. By symmetry, segment NP is a diameter of the circle of O 1 and its midpoint L is the center of O 1. Let line AL intersect circle O again at Z. Let line ZQ intersect line CM at J and circle O again at K.
Since AB and AC are tangent to circle O 1 ,
AL bisects ∠ CAB so that Z is the
midpoint of arc BC. Since Q is the
midpoint of segment BC , ∠ ZQB = 90 ˚ =
∠ LPA and ∠ JQC = 90 ˚ = ∠ MPB. Next
∠ ZBQ = ∠ ZBC = ∠ ZAC = ∠ LAP.
So ∆ ZQB , ∆ LPA are similar. Since M is the midpoint of arc AMB ,
∠ JCQ = ∠ MCB = ∠ MCA = ∠ MBP.
So ∆ JQC , ∆ MPB are similar.
By the intersecting chord theorem, AP·BP = NP·MP = 2 LP·MP. Using the similar triangles above, we have
. 2
BQ CQ
ZQ JQ
AP BP
LP MP
By the intersecting chord theorem, KQ·ZQ = BQ·CQ so that
KQ = ( BQ·CQ ) /ZQ = 2 JQ.
This implies J is the midpoint of KQ. Hence the circle with center J and diameter KQ is tangent to circle O at K and tangent to BC at Q. Since J is on
the bisector of ∠ BCA , this circle is
also tangent to AC. So this circle is O 2.
Example 6. ( 1989 IMO ) In an acute-angled triangle ABC the internal bisector of angle A meets the circumcircle of the triangle again at A 1. Points B 1 and C 1 are defined similarly. Let A 0 be the point of intersection of the line AA 1 with the external bisectors of angles B and C. Points B 0 and C 0 are defined similarly. Prove that:
(i) the area of the triangle A 0 B 0 C 0 is twice the area of the hexagon AC 1 BA 1 CB 1 ,
(ii) the area of the triangle A 0 B 0 C 0 is at least four times the area of the triangle ABC. C 0
B 0 C A 0
B
A I B 1 A 1
C 1
Solution. (i) Let I be the incenter of ∆ ABC. Since internal angle bisector and external angle bisector are
perpendicular, we have ∠ B 0 BA 0 = 90˚.
By the theorem, A 1 I = A 1 B. So A 1 must be the midpoint of the hypotenuse A 0 I of right triangle IBA 0. So the area of ∆ BIA 0 is twice the area of ∆ BIA 1.
Cutting the hexagon AC 1 BA 1 CB 1 into six triangles with common vertex I and applying a similar area fact like the last statement to each of the six triangles, we get the conclusion of (i).
(ii) Using (i), we only need to show the area of hexagon AC 1 BA 1 CB 1 is at least twice the area of ∆ ABC.
B
C
A
A 1
A^ D
2
H
(continued on page 4)
A E B
D
C
F
Let E be the midpoints of AB. Since CD is tangent to the circle, the distance from E to line CD is h 1 = AB /2. Let F be the midpoint of CD and let h 2 be the distance from F to line AB. Observe that the areas of ∆ CEF and ∆ DEF = CD·AB /8. Now
line AB is tangent to the circle with side CD as diameter ⇔ h 2 = CD / ⇔ areas of ∆ AEF , ∆ BEF , ∆ CEF and ∆ DEF are equal to AB·CD / ⇔ AD ∥ EF , BC ∥ EF ⇔ AD ∥ BC.
Problem 249. For a positive integer n ,
if a 1 ,⋯, an , b 1 , ⋯, bn are in [1,2] and
then prove
that
a 1^2 + L+ an^2 = b 12 +L+ b^2 n ,
1
3
1
3 1 n n
n (^) a a b
a b
a
Solution. Jeff CHEN (Virginia, USA).
For x , y in [1,2], we have
1/2 ≤ x / y ≤ 2 ⇔ y /2 ≤ x ≤ 2 y ⇔ ( y /2 − x )(2 y − x ) ≤ 0 ⇔ x^2 + y^2 ≤ 5 xy /2.
Let x = ai and y = bi , then ai^2 + bi^2 ≤ 5 aibi /2. Summing and manipulating, we get
. 5
1
2 2 1
2 1
= = =
n
i
i i
n
i
i
n
i
ai bi a b a
Let x = ( ai^3 / bi )1/2^ and y = ( aibi )1/2. Then x/y = ai/bi in [1,2]. So ai^3 / bi + aibi ≤ 5 ai^2 /2. Summing, we get
1
2 1 1
3
= = =
n
i
i
n
i
i i
n
i (^) i
i (^) ab a b
a
Adding the two displayed inequalities,
we get
1
3
1
3 1 n n
n (^) a a b
a b
a
Problem 250. Prove that every region with a convex polygon boundary cannot be dissected into finitely many regions with nonconvex quadrilateral boundaries. Solution. YUNG Fai.
Assume the contrary that there is a dissection of the region into nonconvex quadrilateral R 1 , R 2 , …, Rn. For a nonconvex quadrilateral Ri , there is a vertex where the angle is θi > 180˚, which we refer to as the large vertex of the quadrilateral. The three other vertices, where the angles are less than 180˚ will be referred to as small vertices.
Since the boundary of the region is a convex polygon, all the large vertices are in the interior of the region. At a large vertex, one angle is θi > 180˚, while the remaining angles are angles of small vertices of some of the quadrilaterals and add up to 360˚ − θi. Now
=
n i i 1
( 360 o θ)
accounts for all the angles associated with all the small vertices. This is a contradiction since this will leave no more angles from the quadrilaterals to form the angles of the region.
Olympiad Corner
(continued from page 1)
Part 2, Day 1 (June 8, 2005)
Problem 1. Determine all triples of positive integers ( a , b , c ), such that a + b + c is the least common multiple of a , b and c.
Problem 2. Let a , b , c , d be positive real numbers. Prove
abcd a^3 b^3 c^3 d^3
a b c d ≤ + + +
Problem 3. In an acute-angled triangle ABC , circle k 1 with diameter AC and k 2 with diameter BC are drawn. Let E be the foot of B on AC and F be the foot of A on BC. Furthermore, let L and N be the points in which the line BE intersects with k 1 (with L lying on the segment BE ) and K and M be the points in which the line AF intersects with k 2 (with K on the segment AF ). Prove that KLMN is a cyclic quadrilateral.
Part 2, Day 2 (June 9, 2005)
Problem 4. The function f is defined for all integers {0, 1, 2, …, 2005}, assuming non-negative integer values in each case. Furthermore, the following conditions are fulfilled for all values of x for which the function is defined: f (2 x + 1) = f (2 x ), f (3 x + 1) = f (3 x ) and f (5 x + 1) = f (5 x ). How many different values can the function assume at most? Problem 5. Determine all sextuples ( a , b , c , d , e , f ) of real numbers, such that the following system of equations is fulfilled: 4 a =( b+c+d+e )^4 , 4 b =( c+d+e+f )^4 , 4 c =( d+e+f+a )^4 , 4 d =( e+f+a+b )^4 , 4 e =( f+a+b+c )^4 , 4 f =( a+b+c+d )^4. Problem 6. Let Q be a point in the interior of a cube. Prove that an infinite number of lines passing through Q exists, such that Q is the mid-point of the line-segment joining the two points P and R in which the line and the cube intersect.
Angle Bisectors Bisect Arcs
(continued from page 2)
Let H be the orthocenter of ∆ ABC. Let line AH intersect BC at D and the circumcircle of ∆ ABC again at A 2. Note
∠ A 2 BC = ∠ A 2 AC
= ∠ DAC
= 90 ˚ − ∠ ACD
= ∠ HBC.
Similarly, we have ∠ A 2 CB = ∠ HCB.
Then ∆ BA 2 C ≅ ∆ BHC. Since A 1 is the midpoint of arc BA 1 C , it is at least as far from chord BC as A 2. So the area of ∆ BA 1 C is at least the area of ∆ BA 2 C. Then the area of quadrilateral BA 1 CH is at least twice the area of ∆ BHC.
Cutting hexagon AC 1 BA 1 CB 1 into three quadrilaterals with common vertex H and comparing with cutting ∆ ABC into three triangles with common vertex H in terms of areas, we get the conclusion of (ii).
Remarks. In the solution of (ii), we saw the orthocenter H of ∆ ABC has the property that ∆ BA 2 C ≅ ∆ BHC (hence, also HD = A 2 D ). These are useful facts for problems related to the orthocenters involving the circumcircles.
