Combinatorial Geometry... Or is it Geometrical Combinatorics? Part II

Geometry Level 3

Circle O has radius 10 units. Point P is on radius OQ and OP = 6 units. How many different chords containing P, including the diameter, have integer lengths?


The answer is 8.

Finn Hulse by Finn Hulse , 21, USA

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2 solutions

Shreyas Shastry
Feb 27, 2014

The shortest possible chord is the one perpendicular to line segment OQ. It will have length = 16, which is integral. 1/2 chord = sqrt[10^2 - 6^2] = 8

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Shortest chord (perpendicular to radius OQ) is 16 units, Longest chord = dia. =20 There can't be more than one chord of length 16 & 20, however, of lengths 17, 18 & 19, two chords each can be accommodated, thus there can be 8 chords: 16, 17,17,18,18,19,19,20 !!! :)

Ananth Rangarajan - 7 years, 2 months ago

i think the answer should be 5 as the shortest chord has length 16 and the longest chord(diameter) has length 20 so there are 5 chords of lengt 16, 17, 18, 19, 20 which are possible

Ishan Tarunesh - 7 years, 3 months ago

how come 16

Hina Goyal - 7 years, 3 months ago
Manu Mehta
Aug 26, 2014

a simple solution can be the number theory approach : P is a point on the radius OQ. Now take a chord AB passing through P and meeting the circle at points A and B. Now as per the property of the chord , AP.PB = OP.PQ = 6.4 = 24 The solution now boils down to the number of factors of 24 i.e. 8 hence the answer.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

three turns given for answer... one attempted ......one guessed............ one left out of frustration : (

ashutosh mahapatra - 6 years, 8 months ago

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