Life of a 3-4-5 triangle

Here are some things I noticed about the 3-4-5 triangle

Let triangle ABC satisfy BC = 3, CA = 4 and AB = 5.

1. Radius of incircle r = 1
Let the incircle touch sides BC, CA and AB at D, E and F respectively. Let the incentre be I.345 1.png
BD + DC + CE + EA + AF + FB = 3 + 4 + 5 = 12
But BD = FB, DC = CE, EA = AF (equal tangents)
So 2(FB + DC + AF) = 12
FB + DC + AF = 6
DC = 6 – (AF + FB) = 6 – 5 = 1 = CE
∠DCE = 90, and ∠IDC = ∠IEC = 90 (radius is perpendicular to tangent), so ∠DIE = 90 by angles in a quadrilateral. Also DC = CE so CDIE is a square. So r = ID = CE = 1.
Alternatively, if you know the formula Area = rs, then this gives r immediately, because area and semiperimeter are both 6. However, the lengths of the tangents will become useful to us later. It is easy to show now that BD = FB = 2 and EA = AF = 3.

(Almost as interestingly, the exradii are 2, 3, and 6.)

2. Let O be the circumcentre of ABC. Then ∠AIO = 45 and ∠BIO = 90
345 3.png
Since the triangle is right-angled, O is the midpoint of AB. So the circumradius R is 5/2.
FB = 2 so OF = 1/2. FI = 1 so OI = √5/2 by Pythagoras. (If you know it you could use OI² = R(R – 2r) to get this)
Also, since BF = 2 and FI = 1, by Pythagoras, BI = √5
But since BI² + OI² = 5 + 5/4 = 25/4 = (5/2)² = OB², ∠BIO = 90 by the converse of Pythagoras.
(Alternatively, OI² = OF × OB which means OI is tangent to the circumcircle of BIF by the converse of tangent-secant theorem. But because ∠BFI = 90, BI is diameter of BIF’s circumcircle by the converse of Thales, so ∠BIO is 90 because tangent is perpendicular to diameter)
Also ∠ABC + ∠BAC = 90, so ∠ABI + ∠BAI = 45. So by angles in a triangle, ∠AIB = 135 so  ∠AIO = 45.
There is a ‘converse’; if ∠AIO = 45 and ∠BIO = 90, then BC : CA : AB = 3 : 4 : 5.
We first establish that A and B are on opposite sides of IO.
345-4
If they were on the same side of OI, then ∠AIB would be 45. Then ∠IBA + ∠IAB = 135 so ∠ABC + ∠BAC = 270, which clearly violates ‘angles in a triangle sum to 180’. (In general it is true that the angle at the incentre is always strictly greater than 90)
345 5.png
So they are on opposite sides of OI. Now the proof is essentially as before but in reverse: ∠AIB = 135 so ∠ACB = 90, so O is the midpoint of AB (using the converse of Thales)
Let the inradius be 1 unit.
As before, OI² = OF × OB, because ∠OIB = ∠BFI = 90, so R² – 2Rr = OF × R.
So OF = R – 2r = R – 2
BF = 2
Also BI² = R² – OI² = 2Rr = 2R (from triangle BIO)
r² + BF² = 1 + 4 = BI² (from triangle BFI)
5 = 2R
R = 5/2
AB = 5
BD = BF = 2; AE = AF = 3
Also because ∠BCA = 90, DIEC is a square again so EC = DC = r = 1 which gives BC = 3, CA = 4 (in units)

3. Let DI meet EF at G. Then G is the centroid of ABC.345 6.png
DG is perpendicular to BC.
For any triangle, AG would pass through the midpoint of BC; this is fairly well-known. G is not always the centroid, however. This fact has no converse – the 3-4-5 triangle is not the only triangle with this property.
345 7.png
Let DG cut AB at X and let BC and FE meet at Y. Let Z be the foot of the perpendicular from F to BC.
Clearly BCA is taken to BDX by an enlargement, centre B, scale factor 2/3 (as BD = 2 and BC = 3), so DX = 8/3.
Also BCA is taken to BZF by an enlargement, centre B, scale factor 2/5 (as BF = 2 and BA = 5), so FZ = 8/5. Also BZ = 6/5 so DZ = 3 – 1 – 6/5 = 4/5
Now let us compute DG.
As FZ = 8/5 and CE = 1, YCE is taken to YZF by an enlargement, centre Y, scale factor 8/5.
So FZ/ZY = EC/CY.

8CY = 5CY + 9
CY=3
So the enlargement taking ECY to GDY, centre Y, has scale factor DY/CY = 4/3. So GD = 4/3 × EC = 4/3
DG is half of DX so G is the midpoint of DX.
345 8.png
Now an enlargement, centre B, scale factor BC/BD = 3/2, takes BDX to BCA. So it takes the midpoint of DX to the midpoint of CA. Thus BT is a median of ABC. Also BG : GT = 2 : 1 so G is the centroid as desired.

4. ABC is the only triangle with integer side lengths for which its perimeter is twice its area.
We know that the area A of a triangle with side lengths a, b and c is given by Heron’s formula:

where s is the semiperimeter

Also, drawing the incircle of the triangle,
integer triangles 1.png
we see that if the lengths a, b and c form a triangle, we can substitute a = x + y, b = y + z, c = z + x for some positive real numbers x, y, and z.
This gives us nicer equations:


So if 2A = P, then




Now


So x + y + z is either a positive integer, or 1/2 more than an integer.
If it is 1/2 more than an integer, then, by subtracting integers x + y, y + z, and z + x in turn, we learn that x, y and z are all 1/2 more than an integer.
But then xyz is 1/8 more than an integer, whereas x + y + z is 1/2 more than an integer. This is impossible so x + y + z is an integer. By subtracting integers x + y, y + z, and z + x in turn, we learn that x, y and z are all integers. So x, y and z are all positive integers.
Clearly, xyz becomes much larger than x + y + z as x, y and z increase, so this suggests a size argument to bound the number of cases to check.


x and y + z are positive so yz – 1 is positive. Hence we can multiply by yz – 1:




Assume WLOG that

Now if y = 1, this bound doesn’t do anything for us, but

z – 1 divides into z + 1 which gives z = 2, x = 3 or z = 3, x = 2. (If z > 3 then z – 1 does not divide z + 1.)
Otherwise, we can use the bounding. If y = 2, we need to check z = 2 (does not give an integer x) and z = 3 (gives x = 1).
If y > 2 then there are no more solutions as

So our solutions are (x, y, z) = (3, 1, 2), (2, 1, 3), (1, 2, 3), all of which yield {a, b, c} = {3, 4, 5}.
This does indeed give 2A = P.

 

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