# Math problem: Looking for elegant proof
**URL:** https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450
**Category:** Factual Questions
**Created:** [March 5, 2009, 10:42pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450 "2009-03-05T22:42:39Z")
**Posts on this page:** 16
**Page:** 1
### Author:  [@Polerius](https://boards.straightdope.com/u/Polerius)
#### Post date: [March 5, 2009, 10:42pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/1 "2009-03-05T22:42:39Z")
```auto
.
A \` .
\ ` .
\ ` . C
\ / ` .
\ / ` .
\/ _________ `. D
B
```
Given
- The angle ABC is 90 degrees
- The angle CBD is 30 degrees
- AB = CD
Wanted
- AC
I know the solution (see spoiler below), but I derived it using brute force (trig identities, etc) and I strongly suspect there is an elegant proof (e.g of the sort “draw a circle with A as the center, and with radius AC, …”) where you can get the answer after drawing a few helpful lines and curves.
[spoiler]AC = [sup]3[/sup]√ 2 AB
(i.e. the cube root of 2 times the length of AB)
The fact that it is a cube root (which one rarely gets as an answer in trig questions) is a bit strange. This somehow further enhances my belief that there may be an elegant solution.
[/spoiler]
BTW, are there any online forums specifically for these sorts of questions, i.e. that are math-specific? Searching for “math forum” on Google does not lead to anything useful.
---
### Author:  [@Frylock](https://boards.straightdope.com/u/Frylock)
#### Post date: [March 5, 2009, 11:09pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/2 "2009-03-05T23:09:07Z")
Using some info from your spoilered section, there’s reason to doubt that there can be a a method using just straightedge and compass.
since the cube root of two can’t be constructed using just those devices. Wikipedia tells me you can do it (construct cube root two) if you’re allowed to make two marks on the straightedge, though.
From your post, though, I gather you wouldn’t insist that only straightedge and compass be used. You can still have an elegant proof using other methods, or even just using some simple modifications of the straightedge/compass thing.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [March 5, 2009, 11:13pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/3 "2009-03-05T23:13:42Z")
The straightedge and compass rule is a limitation on doing constructions, not writing proofs. However, the fact that triangles ABC and BCD aren’t similar makes me suspect that no synthetic proof would really be elegant.
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### Author:  [@Polerius](https://boards.straightdope.com/u/Polerius)
#### Post date: [March 5, 2009, 11:19pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/4 "2009-03-05T23:19:23Z")
BTW, this was a Math Olympiad-ish type question from many years ago, so that is one reason I thought the brute force solution would not be the only one, since these guys were usually not looking for the brute force approach.
---
### Author:  [@Quartz](https://boards.straightdope.com/u/Quartz)
#### Post date: [March 5, 2009, 11:33pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/5 "2009-03-05T23:33:19Z")
Is AD a straight line? Are we looking at a triangle or quadrilateral?
---
### Author:  [@Polerius](https://boards.straightdope.com/u/Polerius)
#### Post date: [March 5, 2009, 11:35pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/6 "2009-03-05T23:35:17Z")
> [@Quartz](#):
>
> Is AD a straight line? Are we looking at a triangle or quadrilateral?
AD is a straight line.
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### Author:  [@SCSimmons](https://boards.straightdope.com/u/SCSimmons)
#### Post date: [March 5, 2009, 11:50pm UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/7 "2009-03-05T23:50:36Z")
I don’t think your solution is correct … try extending BC to E, and AB to F, such that BEDF is a rectangle. _Now_ you have some similar triangles to work with. 🙂
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [March 6, 2009, 12:12am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/8 "2009-03-06T00:12:05Z")
Why don’t you tell us what your (non-elegant) reasoning was, as a start?
---
### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [March 6, 2009, 12:14am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/9 "2009-03-06T00:14:57Z")
A little trigonometry leads to a short proof. In general, if ABD is a triangle and C is a point on the segment AD, sin(ABC)/sin(CBD) = AC/CD. Here ABC = 90[sup]o[/sup] and CBD = 30[sup]o[/sup], so AC = 2CD. From there you can use the law of cosines to get some quadratics where the algebra works out OK, and if I haven’t screwed up my math, AC = 4AB/sqrt(13).
---
### Author:  [@Polerius](https://boards.straightdope.com/u/Polerius)
#### Post date: [March 6, 2009, 12:22am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/10 "2009-03-06T00:22:55Z")
> [@ultrafilter](#):
>
> A little trigonometry leads to a short proof. In general, if ABD is a triangle and C is a point on the segment AD, sin(ABC)/sin(CBD) = AC/CD. Here ABC = 90[sup]o[/sup] and CBD = 30[sup]o[/sup], so AC = 2CD.
I’m not following this.
If that were the case, then AC = 2 AB, since CD=AB.
But that contradicts your finding below that AC = (4/sqrt(13)) AB
> [@](#):
>
> From there you can use the law of cosines to get some quadratics where the algebra works out OK, and if I haven’t screwed up my math, AC = 4AB/sqrt(13).
---
### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [March 6, 2009, 12:27am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/11 "2009-03-06T00:27:48Z")
My mistake. I was using AB = BD.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [March 6, 2009, 12:27am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/12 "2009-03-06T00:27:59Z")
Edited: Nevermind, already addressed immediately above
---
### Author:  [@lobotomyboy63](https://boards.straightdope.com/u/lobotomyboy63)
#### Post date: [March 6, 2009, 12:28am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/13 "2009-03-06T00:28:18Z")
They’re similar triangles, right? If AB = CD and BC=BC (it forms part of each), then the unknown sides must be proportionate. Geometry was long ago…
ETA: Nah that can’t be it. Isoceles and equilaterals, for instance, could have two sides of length A and not be similar.
---
### Author:  [@SCSimmons](https://boards.straightdope.com/u/SCSimmons)
#### Post date: [March 6, 2009, 2:12am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/14 "2009-03-06T02:12:02Z")
> [@lobotomyboy63](#):
>
> They’re similar triangles, right? If AB = CD and BC=BC (it forms part of each), then the unknown sides must be proportionate. Geometry was long ago…
>
> ETA: Nah that can’t be it. Isoceles and equilaterals, for instance, could have two sides of length A and not be similar.
They’re clearly _not_ similar. Angle ABC is right, and angle BCD is necessarily obtuse. (It’s the complement of BCA, which is necessarily acute, as one of the two non-right angles of right triangle ABC.)
---
### Author:  [@Topologist](https://boards.straightdope.com/u/Topologist)
#### Post date: [March 6, 2009, 4:29am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/15 "2009-03-06T04:29:01Z")
Here’s a more elegant proof – at least it doesn’t use any trig.
I won’t attempt to draw the figure, but recommend that you do to follow along. Let E be the point on BD such that CE is perpendicular to BC. Then BCE is a 30-60 right triangle and ABD is similar to CED. From that similarity we get
CD/CE = (AC + CD)/AB;
substituting AB for CD, we get
AB/CE = (AC + AB)/AB.
From the 30-60 right triangle BCE and the right triangle ABC we get
CE = BC/√3 = √(AC[sup]2[/sup] - AB[sup]2[/sup])/√3.
Now substitute that for CE in the second equation above and, to simplify things, write x = AC/AB. We get
√3/√(x[sup]2[/sup] - 1) = x + 1,
which, on squaring and rearranging, gives the quartic equation
x[sup]4[/sup] + 2x[sup]3[/sup] - 2x - 4 = 0,
which has [sup]3[/sup]√2 as its only positive solution. Hence, AC = [sup]3[/sup]√2 AB as **Polerius** got.
---
### Author:  [@Polerius](https://boards.straightdope.com/u/Polerius)
#### Post date: [March 6, 2009, 7:23am UTC](https://boards.straightdope.com/t/math-problem-looking-for-elegant-proof/488450/16 "2009-03-06T07:23:23Z")
> [@Topologist](#):
>
> Here’s a more elegant proof – at least it doesn’t use any trig.
>
> I won’t attempt to draw the figure, but recommend that you do to follow along. Let E be the point on BD such that CE is perpendicular to BC. Then BCE is a 30-60 right triangle and ABD is similar to CED. From that similarity we get
>
> CD/CE = (AC + CD)/AB;
>
> substituting AB for CD, we get
>
> AB/CE = (AC + AB)/AB.
>
> From the 30-60 right triangle BCE and the right triangle ABC we get
>
> CE = BC/√3 = √(AC[sup]2[/sup] - AB[sup]2[/sup])/√3.
>
> Now substitute that for CE in the second equation above and, to simplify things, write x = AC/AB. We get
>
> √3/√(x[sup]2[/sup] - 1) = x + 1,
>
> which, on squaring and rearranging, gives the quartic equation
>
> x[sup]4[/sup] + 2x[sup]3[/sup] - 2x - 4 = 0,
>
> which has [sup]3[/sup]√2 as its only positive solution. Hence, AC = [sup]3[/sup]√2 AB as **Polerius** got.
Awesome. Leave it to a **Topologist** found out the best place to draw a line that helps solve the problem with minimal trigonometry 🙂
Since you posted the above, I might as well post the best one I had come up with
```auto
Let phi = angle BAC
(1) From triangle ABC: BC = AC*sin(phi)
(2) From triangle BCD: sin(60-phi)/BC = sin(30)/CD
==> BC = 2*sin(60-phi) AB
(3) From (1) & (2): AC*sin(phi) = 2*sin(60-phi) AB
==> AC = AB 2*sin(60-phi)/sin(phi)
(4) From triangle ABC: AC = AB / cos(phi)
(5) From (3) & (4) : 2*sin(60-phi)/sin(phi) = 1/cos(phi)
We just need to use the above to find cos(phi), since AC = AB / cos(phi)
2*(sin(60)*cos(phi)-cos(60)*sin(phi)) = sin(phi) /cos(phi)
sqrt(3)*cos(phi)-sin(phi) = sin(phi)/cos(phi)
sqrt(3)*cos(phi)^2 = sin(phi)*(1+cos(phi))
sqrt(3)*cos(phi)^2 = sqrt(1-cos(phi)^2)*(1+cos(phi))
Let x = cos(phi)
sqrt(3)*x^2 = sqrt(1-x^2)*(1+x)
3*x^4 = (1-x^2)*(1+x)^2
4*x^4 + 2*x^3 -2*x - 1 = 0
(2*x + 1)*(2*x^3-1) = 0
The only positive real root of the above equation is x = (1/2)^(1/3)
So, using (4), we have AC = AB*2^(1/3)
```