The Ink- Spot

Geometry Level 3

One day, mom set a very interesting problem to me. She pushed a large circular table we have at home, into the corner of the room, so that it touched both walls and spilled a spot of ink on the extreme edge, and said, " rocky here is a little puzzle for you. Look at that spot. it is exactly eight inches from one wall and nine inches from the other. Now tell me the diameter of the table without measuring it........." Can you?

58 24 64 18

Vishwa Tej by Vishwa Tej , 22, India

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2 solutions

Vishwa Tej
Jul 21, 2015

Double the product of the 2 distances from the wall and you get 144, which is the square of 12. The sum of the 2 distances is 17, and when we add these two no's, 12 and 17 together and also subtract one from the other, we get two answers 29 and 5 as the radius, or half diameter of the table. naturally the diameter should be 58 or 10. However, a table of the latter dimensions cannot be a large circular table and therefore, the table must be 58 in diameter.

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I didn't quite understand the question let alone the answer. Can you please explain the question in more detail? (With a diagram if possible)

Rahul Badenkal - 5 years, 10 months ago

There is a second answer. The diameter could be 10. (It is not one of the options, but still....)

Marta Reece - 3 years, 10 months ago
Tina Sobo
Nov 23, 2016

This question would greatly benefit from a diagram - but as I understood it, there is a large table pushed up into the corner of a room. There's a spot on the edge of the table --> Construction: The table is a circle with center at C, and radius of length r. draw two tangents that meet to at point A to form a right angle (the corner of the room). The tangent points of the wall/table call B and D. Let S be the spot on the circumference of circle C. It is 8 ft from AB and 9 ft from AD. It is therefore r-8 ft from CD and r-9 ft from BC. A radius, CS, is clearly r ft in length.

Triangle SBC is a right triangle, with side lengths r-8, r-9 and hypotenuse r. By the Pythagorean theorem: (r-8)^2 + (r-9)^2 = r^2, solving for r, we get r = 5, 29 and diameter is equal to 10, 58.

If the table were 10 ft in diameter, that would put the spot not within the arc between the two tangent points (closest to the corner). So the diameter must be 58 ft, which is a heck of a table.

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