Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

When I taught myself algebra 2 I got stuck on some 3d geometry problems, I decided I'm come back to them later. Well yesterday I flipped back to them and was able to solve all but one problem.

See pic below.

Given plane k is perpendicular to line AB, and line BE is perpendicular to line CD, outline a proof to show ΔCAD is isosceles.

I worked on this one a while to no avail. To me it seems that point D could lie ANYWHERE along the line that passes through CE, and the perpendicular stipulations would still be met, but triangle CAD would definitly not be isosceles.

*Last edited by mikau (2006-05-12 08:15:14)*

A logarithm is just a misspelled algorithm.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

Thats what it looks like to me - because we could move C or D along that line and all given conditions would be met - but that could result in CE ≠ ED, which would mean that AC ≠ AD

Maybe a mistake in the problem, or in my brain.

Nice illustration, though.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Nice illustration, though.

Thanks!

because we could move C or D along that line and all given conditions would be met - but that could result in CE ≠ ED, which would mean that AC ≠ AD

Yeah exactly thats exactly what I thought. I wonder if the answer is "it cannot be proved." Unfortunatly, saxon does not provide solutions to geometry proofs in the answer key, instead they tell you to refer to the lesson where it was taught.

The symbol for perpendicular is an upsidedown T, right? Or does that mean perpendicular bisector? That would do it.

*Last edited by mikau (2006-05-12 12:14:08)*

A logarithm is just a misspelled algorithm.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

you should give another condition - two angles equal, or two line segments equal.

**X'(y-Xβ)=0**

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I didn't make up this problem, its in a book. But yeah it seems we need one more piece of information.

A logarithm is just a misspelled algorithm.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

The symbol for perpendicular is an upsidedown T, right? Or does that mean perpendicular bisector? That would do it.

Right, it just means perpendicular.

I agree with mathisfun, I believe this is an impossible proof. Now if you were given that BE bisected CD, then it would be possible.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

My rather short page on: Symbols in Geometry

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

If two objects' projecture on a surface are equal, while their angles against the surface are equal, they are equal, too.

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

mikau wrote:

I didn't make up this problem, its in a book. But yeah it seems we need one more piece of information.

Maybe I should have used passive format.

**X'(y-Xβ)=0**

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

George,Y wrote:

Maybe I should have used passive format.

"They" instead of "you" is what you meant, perhaps?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

kylekatarn wrote:

Maybe using a non-standard notion of norm and distance over R² this is true....

Or maybe not...

:P

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I already figured this one out but lets see if anyone else has the same trouble I had. My geometry is a little rusty so you guys will probably get it right away.

Its just that I made an interesting observation with this problem that helped me solve it and I want to see if anyone can solve it without that.

Oh and btw, lines xy and ab are NOT tangent lines to circle p. Well I suppose they could be if angles xyp and abp are right angles, but it is not a given fact in the problem.

*Last edited by mikau (2006-05-17 05:57:48)*

A logarithm is just a misspelled algorithm.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I see only two interesting things, it looks like a cat, and the triangles are equal due to the SSA theorm.

Is that a theorm?? I thought if you have SSS, SSA, or SAA, then they are equal triangles, but I can not remember for sure.

And SAS was one of them too. (Hate it when swears get in the way of math)

*Last edited by John E. Franklin (2006-05-17 09:19:19)*

**igloo** **myrtilles** **fourmis**

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

lol! No A (cough) SS is not a theorem from what I recall there is SSS, AAAS, SAS, and HL, but not SSA.

And not to be picky but I believe they are called congruency postulates, not theorems. I don't think they can be proved formally, I'm not sure.

A logarithm is just a misspelled algorithm.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

I assume you are assigning the fault to the assembler of the swear list? I can assure you that I, with the assistance of associates, was most assiduous!

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

hahahaa...

A logarithm is just a misspelled algorithm.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

1. They ARE theorems.

2. There's a congurency theorem stating that:

If two sides and the opposite angle of the greater of them are equal for 2 triangles then the triangles are congurent. Now, x and a are outside of P , so xp > yp and ap>bp. And yp and bp are radiuses of P, so yp=bp. Then Δxyp = Δabp.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Well I just rechecked my book and it says "we postulate these are true" for SAS, SSS and all those. Perhaps they meant "for now" and didn't bother to do a formal proof. Thanks for the heads up.

Yes krassi, my book never mentioned that theorem to me, thats the same congruency theorem I came up with. I called it LSA. Long, Short, angle. If they appear in that oder in two triangles then they are congruent. Basicly is L is longer then S then there is only one point L intersects the angle formed by A. If L were shorter it would intersect in two points.

This in a way gives birth to the HL theorem, because the hypotenuse will always be longer then a leg, then comes a short side, and then a right triangle.

A logarithm is just a misspelled algorithm.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I know the first three "postulates" as I, II and III congurency indicators.

In the 7th grade bulgarian geometry textbook there are proofs of them.

The LSA is known as IV congurency indicator. There's a proof in 8th grade texbook for it.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**infinit****Guest**

the first problem is proveable!

cuz AB is perp to k

and BC intersects BD at BC ( and they are both in k)

so AB is perp to both BD and BC

so ABC is directly similar to ABD

so AC=AD

so its an isoscles

capishe?

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

could you state your congruency postulates/theorems there, infinit?

"so ABC is directly similar to ABD" by what reasoning? What simularity theorem are you using?

A logarithm is just a misspelled algorithm.

Offline

**orcared****Member**- Registered: 2008-07-12
- Posts: 6

The proposed theorem holds only if plane k is perpendicular to line AB, And BE is a perpendicular bisector of CD.

http://www.idealmath.com

Offline

**qcao****Guest**

Regarding the original 3D problem, it can be easily proven that triangle ACD is isosceles if and only if BE is the perpendicular bisector of CD. Meaning triangle ACD is not isosceles if E is not the midpoint of CD.

Regarding the second problem in quote #13, let's extend xy and ab; they intersects at O. Triangles Oyb is isosceles because angles Oyb and Oby are equal ==> Op is both median and angle bisector. From p, draw pH and pK perpendicular to Ox and Oa, respectively ==> pH = pK. Triangles OHx and OKa are congruent because px = pa, pH = pK, and Hx = Ka = sqrt(px^2 + pH^2). Therefore angle yxp = angle pab. Given that angle xyp = angle abp, the third angle in triangles xyp and abp must be equal ==> angle xpy = angle apb. Finally, triangles xyp and apb are congruent (SAS).

**qcao****Guest**

There is an error in the proof in quote #21: ABC is "directly similar" to ABD.

Both triangles have AB in common, a 90 degree angle, and that's it. They are not similar nor congruent. Actually, it can be easily seen that if ED < EC, then BD is < BC.

**qcao****Guest**

There are two typo errors in quote #24.

Triangles OHx and OKa should read PHx and PKa. Sorry!

Pages: **1**