Let there be a triangle A B C with the area of 10 and the radius of its circumcircle of 5. A point M lies inside the circumcircle of A B C . Suppose a triangle whose side lengths are A M . B C , B M . A C , C M . A B be called Δ ( M ) . Determine the maximum value of the area of Δ ( M ) .
Note: A pedal triangle is obtained by projecting a point onto the sides of a triangle.
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Bad news to this question. I have already found some examples which exceeded 250. I hope I am not making mistake to this question. You can try to verify if what I tell is true.
Your solution is wrong dear. The max area is not 250. But something else.
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I know this problem is controversial but please provide some proof, at least a more logical one than this solution.
The author is introducing a genuine way but I am introducing a conceptual way.
Two isosceles triangles are found for our calculations although right angle triangle and also triangle of various acute angles restricted to certain ranges are also found. For optimization, we prefer equilateral triangle but R = 5 and △ = 10 do not allow for a validity. (24, 35, 121) of integer sides is the closest possible but unable to give exact area of 10. Criteria to satisfy is sin 2 A + sin 2 B + sin 2 C = 2 1 5 2 1 0 = 5 4 .
Formulas applied: D = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 and △ = s ( s − a ) ( s − b ) ( s − c ) = 2 1 ∣ 4 points ∣ .
With ABC of area of 10 with coordinates shown as follows, y-ordinate of M can be varied for obtaining a maximum area wanted. Initially, I put 3 2 of y-ordinate of A or B but there valid with a turning point for maximum with 5 as y-ordinate:
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It is a replacement of equilateral triangle with isosceles triangle for finding such maximum. Otherwise, the span of location of M (x, y) is complicated to simulate even with random generator to be applied.
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This is the case wanted. Its area is 250 (from data onto an area of only 10). Think it as an easy question can make a difficult question to become easy. Nevertheless, I am applying a reflective reliance instead of absolute way to answer this question. You may need to do further determination from my working described. [But my intuition makes me a correct answer to answer this question.] Don't ask me why. [(Ought to take back this sentence.)]
We are told that 1 0 × 5 2 = 250. More convincing to confirm and to remember for typical application:
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For the right angled triangle above, M required is located at edge of half way along the hypotenuse.
Answer: > 2 5 0
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This is a first report to this question. Values > 250 as shown are N O T maximized.
I will ask my lecturer about this. This problem is confusing let alone his solution and your solution @Lu Chee Ket
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And I have checked his solution very carefully before deciding to transfer it into Words and I found it very geometrically correct except for some steps skipped (maybe for who understand pedal triangles)
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You may take
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as an immediate example to check. Coordinates of
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suppose to be fixed, let say for a convention taken.
M (5.179, 0.001) to stay within shall fix for:
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The answer shall reveal immediately.
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Credits go to my lecturer for the problem, solution and the graph