Spieker Table Proofs

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ABC is a 3-4-5 triangle, with midpoints DEF.
Prove that the incenter of triangle ABC is on the incircle of DEF.

This proof doesn't make use of most of the lines on Theo's table, but it
works.

Construct a line parallel to AC through the incenter, and another parallel
to BC through the incenter. You now have a small triangle, similar to ABC,
with right angle at ABC's incenter and hypotenuse along the original
hypotenuse of ABC. The altitude of triangle ABC from the right angle to the
hypotenuse is 12/5. The altitude of the new small triangle is the inradius
of ABC. Using the similar triangles, you can calculate the inradius, which
is 1.

DEF is similar to ABC, with proportion 1/2. DEF's inradius is 1/2.

If you lay a coordinate grid over the table with axes on AC and BC, the
incenter of ABC is at (1,1). The right angle in DEF is at (2, 3/2) and its
incenter is therefore at (3/2, 1).

The distance between the incenters is 1/2, the same as DEF's inradius, so
ABC's incenter must lie on DEF's incircle.

This also shows that the line through the incenters is parallel to one of
the legs of the triangle, which is apparent on diagrams on the Spieker
Circle page on Mathworld, though less so on your page because the angle
bisectors and cleavers create a bit of an optical illusion.

Joseph DeVincentis
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Dear Ed,
I was intrigued by your problem on the Spieker
circle, and solved the general problem: when does the
incentre lie on the Spieker circle?
It is not hard to see that this occurs iff |GI| =
r/3, where G is the centroid, I is the incentre, and r
is the inradius. Crunching through the algebra, this
becomes
(a + b - 3c)*(a + c - 3b)*(b + c - 3a) = 0,
or in other words, the sum of two sides is equal to
three times the third. In the case of the 3-4-5
triangle, 4 + 5 = 3*3.

Cheers,

Naoki Sato
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Let ABC be a 3-4-5 triangle (for convenience, AB = 4, A is the right angle)
with incenter I. Drop perpendicular lines, IX, IY, and IZ from I to AB, BC,
and AC respectively. By definition of incenter, IX = IY = IZ, and since AIX
is a 45-45-90 triangle, AX = IX. Further, since AI, BI, and CI bisect A, B
and C respectively, AX = AZ, CZ = CY, and BX = BY. And finally AX + XB = AB
= 4 etc, so solving for AX, we get AX = 1. Place A at the origin (0,0) with
AB and AC along the positive axes. I must lie at (1,1). Now, draw in DEF.
D as the midpoint of BC, E as the midpoint of AC, F as the midpoint of AB.
The coordinate of D is (1.5, 2). Let H be the incenter of DEF. Since DEF
is similar to ABC, H must be in the same location relative and proportional
to D as I is from A. In other words, D-H = (A - I)/2. this puts H at (1,
1.5). Further more, the radius of the incircle of DEF must be half the
radius of the incircle of ABC. The incircle of ABC has radius IX = 1,
making the incircle of DEF have radius 1/2. So, all points at a distance of
1/2 from H must lie on the incircle of DEF. Coincidently, I has a distance
of 1/2 from H.
David Stigant
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Ed,
I solved your puzzle based on Theo Gray's table.
Since I'm a physicist, not a mathematician, my proof
is straightforward and rather inelegant, but it does
the job:
We will assign a coordinate basis (x, y) to our
triangle, and then locate the points in question.
We seek to prove the incenter of ABC lies on the
incircle of DEF; this is equivalent to showing that
the distance between the incenters of ABC and DEF
is equal to the radius of the incircle of DEF.
Assign our coordinate system such that point A
is at the origin, point B is at (0, 3), and point C
is at (4, 0). The bisector of angle A is
the line y = x. The bisector of angle B can be
found to be y = -2x + 3, since we know it passes
through point B at (0, 3), and we can find its
slope with -tan[90 deg - arccos(3/5)]. These
two bisector lines intersect at the point (1, 1),
which must be the location of the incenter of 
triangle ABC.
The midpoints D, E, F lie at (0, 1.5), (2, 1.5),
and (2, 0) respectively. The bisector of angle E
clearly has slope 1; its equation can be found
to be y = x - 0.5. The bisector of angle F has
slope -2, and passes through (2,0), so its equation
is y = -2x + 4. These two bisectors intersect at
(1.5, 1), which is the location of the incenter of
triangle DEF.
The incircle of DEF must pass through the point
(1.5, 1.5), since the incenter is equidistant
from all 3 sides, the shortest distance between
a point and a line is along a perpendicular to
the line, and this perpendicular to side DE is
a vertical line passing through the incenter at
x=1.5. So, the radius of the incircle of DEF is
the distance between (1.5, 1) and (1.5, 1.5),
which is 0.5.
The distance between the two incenters of our
triangles is the distance between (1, 1) and
(1.5, 1), which equals 0.5. Since the incenter
of ABC is 0.5 away from the incenter of DEF,
and the incircle of DEF is a circle of radius
0.5 centered at the incenter of DEF, the
incenter of ABC must lie on the incircle of
DEF.
This is probably not the most rigorous proof
you'll receive, but I don't think I skipped over
proving anything that's not obvious.
Happy puzzling!

-Dan Hennessy

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Hi Ed,

Here's a proof using Cartesian geometry, that the incenter of triangle ABC is on the incircle of DEF:

Let A be at (0,4); B at (0,0) and C at (3,0).
Let O = ABC incenter.

Therefore BO follows line y = x [ABO = 90°/2 = 45°]

AO follows line y = 4 - x/Tan(a) where 2a = <BAC; Tan(2a) = BC/BA = 3/4

Using the double angle formula: Tan(2a) = 2.Tan(a) / (1 - Tan²(a)) = 3/4
which gives 2 solutions, only one of which is possible by inspection: Tan(a) = 1/3.

So AO follows y = 4 - 3x; BO follows y = x
and O is where the two lines intersect, at (1,1).

Now let D = mid-point of AB at (0,2)
E = mid-point of AC at (1.5,2)
F = mid-point of BC at (1.5,0)
P = incenter of DEF.
FED and ABC are similar triangles; EF = AB/2; DE = BC/2; DF = AC/2

Bisecting <DEF: DEP = 45°, so EP must follow y = x + 0.5 [gradiant = 1, passes through E]
DP follows y = 2 - x.Tan(b) where Tan(2b) = EF/DE = 2/1.5 = 4/3

Using the double angle formula: Tan(2b) = 2.Tan(b) / (1 - Tan²(b)) = 4/3
which again gives 2 solutions, only one of which is possible by inspection: Tan(b) = 1/2.

So DP follows y = 2 - x/2
EP follows y = x + 0.5
and P is where the two lines intersect, at (1,1.5).

Now, DE is horizontal along y = 2 and by definition is tangential to DEF incircle
Therefore DEF incircle radius must be the difference in the y coordinates = 2 - 1.5 = 0.5
So DEF incircle is described by (x - 1)² + (y - 1.5)² = 0.5²

O, the ABC incenter is at (1,1) [as shown earlier]
Therefore, substituting for x & y: (0)² + (-0.5)² = 0.5² which is true

Therefore the ABC incenter lies on the DEF incircle. qe.d.!

Best wishes, and a happy and peaceful Christmas and New Year to you.

Hugh Rutherford.-