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.-