## Two difficult geometry problems

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jaap
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### Two difficult geometry problems

I have two tricky geometry problems begging for proofs. I have tried and failed to find classic geometry proofs for these.

1) A quadrangle ABCD satisfies the following relation on the angles:
2A + C = 360 degrees
and the following on the edge lengths:
BC = CD = DA

Prove that angle D is 60 degrees.

It is easy to see that given the three equal side lengths, and D=60, then it follows that 2A+C=360. If you change angle D then the angle 2A+C also changes and so can no longer be 360. However I don't know how to prove the opposite direction in a nice way without using trig.

I'm doing a bit of research into tiling patterns, and the above quadrangle makes quite a nice tiling, and it surprised me when I noticed that the requirements imposed on the tile by the tiling forced one angle to be constant. Here is a picture of the tiling pattern using this tile, with two different choices of side length AB.

Here is a second, and I think more difficult one I ran into.

2) A quadrangle ABCD satisfies the following relation on the angles:
A + C = 180 degrees
and the following on the edge lengths:
AB = DA
BC = CD + DA

Prove that D is 120 degrees.

I have no clue how to prove this classically, and even using trig to prove it seems to get unwieldy quickly. Is there a relatively simple proof of this?

eta oin shrdlu
Posts: 442
Joined: Sat Jan 19, 2008 4:25 am UTC

### Re: Two difficult geometry problems

For the first one:
Spoiler:
Draw segment CD, and draw the circles of radius CD centered at C and D. B must lie on circle C, and A must lie on circle D. Choose the point B, on circle C outside circle D. The arc BD on circle C has central angle BCD. So the locus of A such that angle BAD=(360°-BCD)/2 is just this arc BD (the inscribed angle is half the intercepted arc). So A must lie at the intersection of the circles C and D, and so ACD is equilateral.

My proof for the second one is kind of ugly right now; I want to see if it will clean up at all.

++\$_
Mo' Money
Posts: 2370
Joined: Thu Nov 01, 2007 4:06 am UTC

### Re: Two difficult geometry problems

For number 2 (with some trig):
Spoiler:
For your quadrilateral, if theta is the angle between CD and DA, then cos theta = (CD2 + DA2 - AB2 - BC2)/2((AB)(BC) + (CD)(DA)). (You just have to use the law of cosines twice. You also have to use the fact that opposite angles sum to 180 once.)

Now it is easy to check that cos theta = -1/2.

A non-trigonometric proof should hopefully be available, though, because the quadrilateral is cyclic....

eta oin shrdlu
Posts: 442
Joined: Sat Jan 19, 2008 4:25 am UTC

### Re: Two difficult geometry problems

OK, here's one for #2 with no trig:
Spoiler:
Because A+C=180°, ABCD is cyclic. Place X on BC with BX=AB=AD, so that CX=CD. Hence triangle CDX is isosceles, and so the angle bisector of angle DCX passes through A. So triangle ADX is also isosceles with AD=AX, and so triangle ABX is equilateral. But B+D=180°, so D=120°.

jaap
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Joined: Fri Jul 06, 2007 7:06 am UTC
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### Re: Two difficult geometry problems

eta oin shrdlu wrote:OK, here's one for #2 with no trig:
Spoiler:
Because A+C=180°, ABCD is cyclic. Place X on BC with BX=AB=AD, so that CX=CD. Hence triangle CDX is isosceles, and so the angle bisector of angle DCX passes through A. So triangle ADX is also isosceles with AD=AX, and so triangle ABX is equilateral. But B+D=180°, so D=120°.

Spoiler:
"Hence triangle CDX is isosceles, so the angle bisector of angle DCX passes through A". How does CDX being isosceles say anything about the location of A?

eta oin shrdlu
Posts: 442
Joined: Sat Jan 19, 2008 4:25 am UTC

### Re: Two difficult geometry problems

jaap wrote:I don't follow this step:
Spoiler:
"Hence triangle CDX is isosceles, so the angle bisector of angle DCX passes through A". How does CDX being isosceles say anything about the location of A?

Spoiler:
The angle bisector of the inscribed angle DCB=DCX passes through the center of the arc DB (inscribed and intercepted arcs again), which (since AB=AD) is the point A.

jaap
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Joined: Fri Jul 06, 2007 7:06 am UTC
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### Re: Two difficult geometry problems

eta oin shrdlu wrote:
jaap wrote:I don't follow this step:
Spoiler:
"Hence triangle CDX is isosceles, so the angle bisector of angle DCX passes through A". How does CDX being isosceles say anything about the location of A?

Spoiler:
The angle bisector of the inscribed angle DCB=DCX passes through the center of the arc DB (inscribed and intercepted arcs again), which (since AB=AD) is the point A.

I may be being dense, but I'm still not seeing this.
Spoiler:
When you say arc DB, you mean the arc centred on A?

eta oin shrdlu
Posts: 442
Joined: Sat Jan 19, 2008 4:25 am UTC

### Re: Two difficult geometry problems

jaap wrote:I may be being dense, but I'm still not seeing this.
Spoiler:
When you say arc DB, you mean the arc centred on A?
Spoiler:
Right. The angle bisection gives two equal inscribed angles at C, which therefore intercept equal arcs of the circle. These two arcs together form arc DAB, and since A bisects this arc, the bisector of angle C passes through A.

jaap
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Joined: Fri Jul 06, 2007 7:06 am UTC
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### Re: Two difficult geometry problems

eta oin shrdlu wrote:
jaap wrote:I may be being dense, but I'm still not seeing this.
Spoiler:
When you say arc DB, you mean the arc centred on A?
Spoiler:
Right. The angle bisection gives two equal inscribed angles at C, which therefore intercept equal arcs of the circle. These two arcs together form arc DAB, and since A bisects this arc, the bisector of angle C passes through A.

Spoiler:
Oh you're talking about arcs from the circle that circumscribes the quadrangle! You didn't make that clear.
Let me re-write the proof a bit.

A+C=180, so the quadrangle is cyclic. Draw the circle that goes through the points A,B,C,D.
Since AD=AB, the corresponding arcs on the circle are equal.
Draw CA.
The angles DCA and ACB subtend equal arcs so are equal too.
Draw point X on BC such that CD=CX.
Triangles ACD and ACX are congruent (share 2 sides and an angle)
We're given that BC=CD+AB, so BX = BC-CX = CD+AB-CX = AB = AD = AX
Therefore triangle ABX is regular and angle B=60.
D = 360-A-B-C = 360-180-60 = 120.

It makes sense now. Thanks!

eta oin shrdlu
Posts: 442
Joined: Sat Jan 19, 2008 4:25 am UTC

### Re: Two difficult geometry problems

jaap wrote:
Spoiler:
Oh you're talking about arcs from the circle that circumscribes the quadrangle! You didn't make that clear.
It makes sense now. Thanks!
Oops, I think I edited that sentence out... Glad you were able to decipher it anyway.