IMO2003 problem B1

eepiccolo · 2003-07-24, 12:21

Here is the fourth problem from the 44th International Mathematical Olympiad. The other five are posted also. I don't know the answers, but I'll be working on them when I get the chance.

Quote:

B1. ABCD is cyclic. The feet of the perpendicular from D to the lines AB, BC, CA are P, Q, R respectively. Show that the angle bisectors of ABC and CDA meet on the line AC iff RP = RQ.

hyh1048576 · 2003-07-24, 15:26

Solution:
First, RP=AD*sinA, RQ=CD*sinC. So RP/RQ=(AD/CD)*(sinA/sinC)=(AD/CD)/(AB/BC)......That means RP=RQ iff AB/BC=AD/CD. Suppose the bisector of ABC meets AC at M and the one of ADC at N. They meet at AC iff AM/CM=AN/CN. On the other hand, AB/BC=AM/CM, AD/DC=AN/CN...... :) :) :) OVER

trif · 2003-07-24, 15:37

I think my problem would be not understanding the terminology.

Cyclic?
"Feet of the perpendicular?"

hyh1048576 · 2003-07-24, 15:57

Cyclic means ABCD is a cyclic quadrilateral(Hope it won't confuse you :? )
Feet of the perpendicular......to P that means DP is perpendicular to AB and DP meets AB at P and so does Q,R :)
Understand?

Jwb52z · 2003-07-24, 23:57

I'm too visual a learner for these things. I would need a picture and a dictionary to keep in mind what everything is anymore lol.

Kevin · 2003-07-25, 02:22

Cyclic means the quadrilateral can be inscribed in a circle

ET_ · 2003-07-25, 12:17

Quote:

Cyclic means the quadrilateral can be inscribed in a circle

Are there quadrilaterals that can not? :-O

Luigi

dswanson · 2003-07-25, 14:34

Any quadrilateral with an interior angle greater than 180 degrees at any vertex cannot be inscribed in a circle. Think of a 4-point approximation of a crescent.

ET_ · 2003-07-25, 18:53

I see... I thought the question was made on a close convex quadrilateral.

Luigi

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2003-07-24, 15:26	#2
hyh1048576 Jun 2003 100₈ Posts	Solution: First, RP=ADsinA, RQ=CDsinC. So RP/RQ=(AD/CD)*(sinA/sinC)=(AD/CD)/(AB/BC)......That means RP=RQ iff AB/BC=AD/CD. Suppose the bisector of ABC meets AC at M and the one of ADC at N. They meet at AC iff AM/CM=AN/CN. On the other hand, AB/BC=AM/CM, AD/DC=AN/CN...... :) :) :) OVER

2003-07-24, 15:37	#3
trif Aug 2002 2×101 Posts	I think my problem would be not understanding the terminology. Cyclic? "Feet of the perpendicular?"

2003-07-24, 15:57	#4
hyh1048576 Jun 2003 2⁶ Posts	Cyclic means ABCD is a cyclic quadrilateral(Hope it won't confuse you :? ) Feet of the perpendicular......to P that means DP is perpendicular to AB and DP meets AB at P and so does Q,R :) Understand?

2003-07-24, 23:57	#5
Jwb52z Sep 2002 1436₈ Posts	I'm too visual a learner for these things. I would need a picture and a dictionary to keep in mind what everything is anymore lol.

2003-07-25, 02:22	#6
Kevin Aug 2002 Ann Arbor, MI 433 Posts	Cyclic means the quadrilateral can be inscribed in a circle

2003-07-25, 14:34	#8
dswanson Aug 2002 2³×5² Posts	Any quadrilateral with an interior angle greater than 180 degrees at any vertex cannot be inscribed in a circle. Think of a 4-point approximation of a crescent.

2003-07-25, 18:53	#9
ET_ Banned "Luigi" Aug 2002 Team Italia 3²·5·107 Posts	I see... I thought the question was made on a close convex quadrilateral. Luigi