An interesting problem related to Pedal Triangle

Geometry Level 5

Let there be a triangle A B C ABC with the area of 10 and the radius of its circumcircle of 5. A point M M lies inside the circumcircle of A B C ABC . Suppose a triangle whose side lengths are A M . B C , B M . A C , C M . A B AM.BC, BM.AC, CM.AB be called Δ ( M ) \Delta(M) . Determine the maximum value of the area of Δ ( M ) \Delta(M) .

Note: A pedal triangle is obtained by projecting a point onto the sides of a triangle.


The answer is 250.

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2 solutions

Leah Smith
Nov 29, 2015

Credits go to my lecturer for the problem, solution and the graph

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.

Lu Chee Ket - 5 years, 6 months ago

Your solution is wrong dear. The max area is not 250. But something else.

Reetun Maiti - 5 years, 6 months ago

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I know this problem is controversial but please provide some proof, at least a more logical one than this solution.

Leah Smith - 5 years, 6 months ago
Lu Chee Ket
Dec 1, 2015

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 \bigtriangleup = 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 = 10 1 2 5 2 \sin 2 A + \sin 2 B + \sin 2 C = \frac{10}{\frac12 5^2} = 4 5 . \displaystyle \frac45.

Formulas applied: D = ( x 2 x 1 ) 2 + ( y 2 y 1 ) 2 D = \sqrt{(x_2 -x_1)^2 + (y_2 - y_1)^2} and \bigtriangleup = s ( s a ) ( s b ) ( s c ) \sqrt{s (s - a)(s - b)(s - c)} = 1 2 \frac12 | 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 2 3 \frac23 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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A   -1.01042350456793   9.89684024054458
B   1.01042350456793    9.89684024054458
C   0   0
M   0   5.00000000000000 inside triangle ABC; chosen or preferred. 

    a = 49.7414320273969
    b = 49.7414320273969
    c = 10.1042350456793
    s = 54.7935495502366
    Area =  250


A   -4.24392501263905   2.35630930570602
B   4.24392501263905    2.35630930570602
C   0   0
M   0   5.00000000000000 outside triangle ABC; not chosen or not preferred.

    a = 24.2709152366882
    b = 24.2709152366882
    c = 42.4392501263905
    s = 45.4905402998835
    Area =  250

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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a = 49.7414320273969
b = 49.7414320273969
c = 10.1042350456793

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 10 × 10 \times 5 2 5^2 = 250. More convincing to confirm and to remember for typical application:

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A   10  0
B   9.582575695 2
C   0   0
M   5   0

    a = 48.9453156465352
    b = 50.0000000000000
    c = 10.2154821844610
    s = 54.5803989154981
    Area =  250

For the right angled triangle above, M required is located at edge of half way along the hypotenuse.

Answer: > 250 \boxed{> 250}

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3 examples exceeded 250:

[1]
A = 90                      10.00000000000000   0
B = 78.2109107608991        9.582575695 2
C = 11.7890892391009        0   0
Sum =   0.799999999999999   10.00000000000000   0
Area =  9.99999999999999    9.58257569495584    2
a = 10.00000000000000       0   0
b = 9.78906312930703        10
c = 2.04309643689220        5.18    0
s = 10.9160797830996            
Area =  10          
a = 48.2            
b = 47.33563473         
c = 10.58323954         
s = 53.05943714         
Area =  250.3741202         

[2]
A = 86                      9.97564050259824    0
B = 82.3285707046987        9.705587526 2.004883796
C = 11.6714292953013        0   0
Sum =   0.800000000000000   9.97564050259824    0
Area =  9.99999999999999    9.70558752640135    2.004883796
a = 9.97564050259824        0   0
b = 9.91049889101209        10
c = 2.02298977903180        5   0.019
s = 10.9545645863211            
Area =  10          
a = 49.63556281         
b = 50.61761324         
c = 10.11502192         
s = 55.18409899         
Area =  251.0305482         

[3]
A = 87                      9.98629534754574    0
B = 81.3109739939435        9.680224906 2.002744692
C = 11.6890260060565        0   0
Sum =   0.800000000000003   9.98629534754574    0
Area =  10.00000000000000   9.68022490552900    2.002744692
a = 9.98629534754574        0   0
b = 9.88522840014036        10
c = 2.02599738815080        5.019   0
s = 10.9487605679184            
Area =  10          
a = 49.60487842         
b = 50.15036978         
c = 10.16848089         
s = 54.96186454         
Area =  251.9056644         

This is a first report to this question. Values > 250 as shown are N O T NOT maximized.

I will ask my lecturer about this. This problem is confusing let alone his solution and your solution @Lu Chee Ket

Leah Smith - 5 years, 6 months ago

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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)

Leah Smith - 5 years, 6 months ago

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You may take

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A = 90                      
B = 78.2109107608991
C = 11.7890892391009  

as an immediate example to check. Coordinates of

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A (10, 0)
B (9.58257569495584, 2)
C (0, 0)

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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a' =    48.2100010371292
b' =    47.3405002579782
c' =    10.5811966439128
s' =    53.0658489695101
Area = 250.355480587124

The answer shall reveal immediately.

Lu Chee Ket - 5 years, 6 months ago

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