Area without the required lengths.

Geometry Level 3

Let $P$ be a point inside an equilateral triangle $ABC$ such that $PA = 3, PB =4, PC =5.$ Find the area of $\triangle ABC$ to 2 decimal places.

The answer is 19.82.

3 solutions

Angel Krastev
Jul 9, 2017

The area is 19.82.
If I move P inside, then 3, 4 and 5 will change, while the area stays the same.
If P goes to the center then AP = BP = CP = (3 * 4 * 5)^(1/3) =
= 60^0.333333 ... = 3.9148676411688635954249307174727.
The central angle APC = 120 degrees.
I can't explain right now why is that. It's my intuition leading me.
Now the area = 1/2 * 3 * AP * BP*sin(120).

Same idea as Angel Krastev's method, but if we moved 4 to the center, and said that the sum of the resulting lines is the same as 3+4+5, so 12/3, we have 4 for each line. Then, if we draw a apothem and make a 30 60 90 right triangle, we have 4 for the hypothenuse, and 4 rad 3 for one whole side of the triangle. This gives us 20.78, not 19.82

Joseph Park - 3 years, 11 months ago

Intuition is not a solution! Although it seems you've got the right answer, I won't accept your solution unless you prove (or refer a theorem stating) that r = (3 4 5)^1/3.

Horia Tudosie - 3 years, 11 months ago

Brenda Meshejian
Jul 11, 2017

Amed Lolo
Jul 10, 2017

Put angle APC =t ,and members of ∆abc =k. So cos(t)=(k^2+25-9)÷(2×k×5)=(k^2+16)÷(10k), angle PCB =60-t ,cos(60-t)= (25+k^2-16)÷(10k)=(9+k^2)÷10k cos(60-t)=cos(60)×cos(-t)-sin(60)×sin(-t) so (9+k^2)÷(10k)=.5×(k^2+16)÷10k+(√3÷2)×(100k^2-(k^2+16)^2)^.5÷(10k) so. 18+2k^2=k^2+16+√(3×(100k^2-(k^2+16)^2)) so (k^2+2)=√(300k^2-3(k^2+16)^2) so by squaring two terms 4k^4-200k^2+772=0 so k=2.053141 (rejected) OR 6.76643( right answer) so area of∆ ABC =.5×6.766^2×√3×.5=19.825#######

Area without the required lengths.

The answer is 19.82.

3 solutions

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