A Geometry Problem Revised
by luimichael, Nov 9, 2007, 1:17 PM
Given that ABC is an equilateral triangle of unknown side. P is a point inside triangle ABC with PA = 4, PB = 3 and PC = 5.
Find the size of the equilateral triangle.
Solution:
Click to reveal hidden text
Following the idea in the above solution, suppose the problem is defined in a more general setting as follows:
Given four points P, A,B and C in the same plane with PA = a, PB = b, PC = c and AB =BC =CA = x.
Determine the value of x in terms of a, b and c.
The solution is neat:
,
where
represents the area of triangle with sides a, b and c.
The plus or minus sign is determined by the following rule: Take a plus sign if the point P is inside triangle ABC and of course take a minus sign if P is outside triangle ABC.
Proof:
Click to reveal hidden text
Alternative proof:Click to reveal hidden text
Further questions:
1. What happens if P is outside triangle ABC ?
2. What is the formula for the general case?
That is to say, what is the expression for x if the lengths PA, PB and PC are given to be a, b and c respectively?
3. How comes the solution of this problem is apparently the same as that of the Fermat Point Problem
Answer:
Click to reveal hidden text
Find the size of the equilateral triangle.
Solution:
Click to reveal hidden text
Let AB = BC = CA = x.
Rotate triangle PBC about point B through 60 degrees. Suppose the point P is rotated to Q.
Then triangle CPB and triangle AQB are congruent.
BQ = 3 , AQ = 5 and PQ = x.
Area of quadrilateral APBQ is

by Brestchneider's Formula
On the other hand, area of quadrilateral APBQ
= area of triangle PBQ + area of triangle PAQ
=
{Note that triangle PAQ is a right-angled triangle}
=
Therefore,
Solving the equation,
.
Rotate triangle PBC about point B through 60 degrees. Suppose the point P is rotated to Q.
Then triangle CPB and triangle AQB are congruent.
BQ = 3 , AQ = 5 and PQ = x.
Area of quadrilateral APBQ is

by Brestchneider's Formula
On the other hand, area of quadrilateral APBQ
= area of triangle PBQ + area of triangle PAQ
=

=

Therefore,

Solving the equation,

Following the idea in the above solution, suppose the problem is defined in a more general setting as follows:
Given four points P, A,B and C in the same plane with PA = a, PB = b, PC = c and AB =BC =CA = x.
Determine the value of x in terms of a, b and c.
The solution is neat:

where

The plus or minus sign is determined by the following rule: Take a plus sign if the point P is inside triangle ABC and of course take a minus sign if P is outside triangle ABC.
Proof:
Click to reveal hidden text
Consider the case where P is inside triangle ABC.
http://www.artofproblemsolving.com/Forum/album.php?mode=pic&id=11202
With reference to the figure in the link, area of quadrilateral APCQ
=
.
Sum of areas of triangles PAQ and PCQ =
.
Therefore, we have

Multiply both sides by four and then squaring both sides:


![$ + [2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)]$](//latex.artofproblemsolving.com/3/c/d/3cdf41dd19b9c7bacc9721d25625f009bf7665ab.png)

Hence,

http://www.artofproblemsolving.com/Forum/album.php?mode=pic&id=11202
With reference to the figure in the link, area of quadrilateral APCQ
=

Sum of areas of triangles PAQ and PCQ =

Therefore, we have

Multiply both sides by four and then squaring both sides:


![$ + [2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)]$](http://latex.artofproblemsolving.com/3/c/d/3cdf41dd19b9c7bacc9721d25625f009bf7665ab.png)

Hence,


Alternative proof:Click to reveal hidden text
By Coordinate Geometry
Again let PA = a, PB = b , PB = c, and AB = BC = CA = t.
Regarding B as the origin and BC as the x-axis.
Then coordinates of A =
.
P is the point of intersection of the following three circles:
-
(1)
-
(2)
-
(3)
Expand (2):
-
(4)
Expand (3):
which reduces to
-
(5)
(1) - (4):
-
(6)
(1) -(5):
Therefore,
After simplification:
-
(7)
Plugging (6) and (7) into (1):

After some lengthy calculations we have
-
(*)
Completing square:

![$ (t^2 - \frac {a^2 + b^2 + c^2}2 )^2 = \frac 3 4 [2(a^2b^2 + b^2c^2 + c^2a^2 ) - (a^4 + b^4 + c^4)]$](//latex.artofproblemsolving.com/2/3/1/231b5f0e13d84e22d8c165b260ff06ee13598d82.png)
, where
= Heron(a,b,c).

Hence,
(**)
When P is inside triangle ABC,
, we take a plus sign in (**).
When P is outside triangle ABC,
, and we take a minus sign in (**).
Again let PA = a, PB = b , PB = c, and AB = BC = CA = t.
Regarding B as the origin and BC as the x-axis.
Then coordinates of A =

P is the point of intersection of the following three circles:

(1)

(2)

(3)
Expand (2):

(4)
Expand (3):

which reduces to

(5)
(1) - (4):

(6)
(1) -(5):

Therefore,

After simplification:

(7)
Plugging (6) and (7) into (1):

After some lengthy calculations we have

(*)
Completing square:

![$ (t^2 - \frac {a^2 + b^2 + c^2}2 )^2 = \frac 3 4 [2(a^2b^2 + b^2c^2 + c^2a^2 ) - (a^4 + b^4 + c^4)]$](http://latex.artofproblemsolving.com/2/3/1/231b5f0e13d84e22d8c165b260ff06ee13598d82.png)



Hence,

(**)
When P is inside triangle ABC,

When P is outside triangle ABC,

Further questions:
1. What happens if P is outside triangle ABC ?
2. What is the formula for the general case?
That is to say, what is the expression for x if the lengths PA, PB and PC are given to be a, b and c respectively?
3. How comes the solution of this problem is apparently the same as that of the Fermat Point Problem
Answer:
Click to reveal hidden text
This type of questions can be solved by the volume formula of tetrahedron.
http://mathworld.wolfram.com/Cayley-MengerDeterminant.html
When the tetrahedron is degenerated to a planar object, the volume should be zero.
The problem is solved.
http://mathworld.wolfram.com/Cayley-MengerDeterminant.html
When the tetrahedron is degenerated to a planar object, the volume should be zero.
The problem is solved.