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 June 14th, 2014, 07:02 PM #251 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra So it doesn't. I didn't look far enough down the page in wikiproof. June 14th, 2014, 07:58 PM #252 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Just give me until tomorrow; I almost have it.  June 15th, 2014, 06:04 PM   #253
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 Originally Posted by eddybob123 Just give me until tomorrow; I almost have it.         :h elp:       :hel p:     June 17th, 2014, 12:42 PM #254 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Ok, sorry it took so long, but I still don't have the answer. Here's my work (not that anyone's going to read through the whole thing: Let $x=AD$, then $AF=DF=x$ also, since the triangle is equilateral. This means the radius of the circle is $r=\frac{x\sqrt{3}}{3}$. Now, since $\angle BFD=120^\circ$, we can use the cosine law on $\displaystyle \triangle BDF$and solve for $FB$: $$BD^2=DF^2+FB^2-2\times DF\times FB\times \cos \angle BFD$$ $$16 = x^2 + FB^2 -2x\times FB \times\frac{-1}{2}$$ $$FB^2+xBF-16=0$$ $$FB=\frac{-x+\sqrt{64-3x^2}}{2}$$ Similarly, we can use the cosine law on $\displaystyle \triangle CDF$ to find $\displaystyle CD=\frac{-x+\sqrt{16-3x^2}}{2}$ Because they're tangents, we know that $CD$ and $FB$ are equal to $CE$ and $BE$ respectively, hence, $$BC=CD+FB=-x+\frac{\sqrt{16-3x^2}+\sqrt{64-3x^2}}{2}$$ -------------------------------------------------------------------------- A well known formula for the radius of an incircle is $r=\frac{A}{s}$, where $r$ is the radius, $A$ is the area of the triangle, and $s$ is the semiperimeter. The area of the triangle is $$\frac{\sin 60^\circ}{2}\times AC\times AB = \frac{\sqrt{3}}{4}\bigg(\frac{x+\sqrt{16-3x^2}}{2}\bigg)\bigg(\frac{x+\sqrt{64-3x^2}}{2}\bigg)$$ and the semiperimeter is $AD+BC$, which is $$\frac{\sqrt{64-3x^2}+\sqrt{16-3x^2}}{2}$$ Hence we get the following equation for $x$: $$\frac{\sqrt{3}}{4}\bigg(\frac{x+\sqrt{16-3x^2}}{2}\bigg)\bigg(\frac{x+\sqrt{64-3x^2}}{2}\bigg)=\frac{x\sqrt{3}}{3}\bigg(\frac{\sq rt{ 64-3x^2}+\sqrt{16-3x^2}}{2}\bigg)$$ which, after simplication of course, does not reveal any quadratic roots. June 17th, 2014, 04:58 PM #255 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra I've found a formula that says: If the points at which an incircle is tangent to a triangle divide the sides of said triangle into lengths x, y and z, then the radius of the incircle is $$r = \sqrt \frac{xyz}{x+y+z}$$ It's a quite beautiful result which yields a quick result here. I'd like to find a proof though. June 17th, 2014, 06:59 PM #256 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra Given a triangle ABC with an incircle O being tangent to the triangle at D, E and F. Let $AF = x$, $BD=y$ and $CE=z$. Then DO is the height of triangle BOC, EO is the height of triangle COA and FO is the height of triangle AOB. And DO, EO and FO are all radii of the circle, having length $r$. Thus, the area of the triangle $\triangle{ABC}=\triangle{AOB} + \triangle{BOC} + \triangle{COA}$ and $$\triangle{ABC} = \frac12 AB \cdot r + \frac12 BC \cdot r + \frac12 CA \cdot r = \frac12 ( x+y ) r + \frac12 ( y+z) r + \frac12 (z+x)r = (x + y + z)r = sr$$ where $s = x + y + z$, the semi-perimeter of the triangle. Now, Heron's formula for the area of the triangle with sides a, b and c is $$\triangle{ABC} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{s((x+y+z)-(x+y))((x+y+z)-(y+z))((x+y+z)-(z+x))} = \sqrt{sxyz}$$ Thus we have \begin{align*} sr &= \sqrt{sxyz} \\ r &= \sqrt{\frac{xyz}{s}} = \sqrt{\frac{xyz}{x+y+z}} \\ \end{align*} June 17th, 2014, 07:38 PM #257 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Does this mean that my work is incorrect? Did you check over it? June 17th, 2014, 08:34 PM #258 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra No, because you didn't get a quadratic. Although I can now see the reason why it was a tad difficult. My diagram has points B and C transposed. Both should be along the sides of the triangle. BD should be from the bottom left corner to D. CF is from the bottom right corner to F. Sorry. June 18th, 2014, 12:54 PM #259 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Oh, that's really easy then ... $\displaystyle BC=6$ We can use the formula $A=rs$ with $A=\frac{\sqrt{3}}{4}(x+2)(x+4)$, $r=\frac{x\sqrt{3}}{3}$, and $s=x+6$, to obtain $$\frac{x\sqrt{3}}{3}(x+6)=\frac{\sqrt{3}}{4}(x+2) (x+4)$$ which gives the quadratic $$x^2+6x-24$$ Taking the positive root only, we get the result $x=-3+\sqrt{33}$. June 18th, 2014, 01:58 PM #260 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra Yep. Well done. You have control. 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