MHB Problem of the Week # 254 - Feb 27, 2017

[Ackbach]

Here is this week's POTW:

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Right triangle $ABC$ has right angle at $C$ and $\angle BAC =\theta$; the point $D$ is chosen on $AB$ so that $|AC|=|AD|=1$; the point $E$ is chosen on $BC$ so that $\angle CDE = \theta$. The perpendicular to $BC$ at $E$ meets $AB$ at $F$. Evaluate $\displaystyle\lim_{\theta\rightarrow 0}|EF|$.

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[Ackbach]

Re: Problem Of The Week # 254 - Feb 27, 2017

This was Problem B-1 in the 1999 William Lowell Putnam Mathematical Competition.

Sorry about the late post. It's been a crazy week, what with an 11pm - 7am work shift (highly unusual for me), my daughter's cochlear implant surgery, and Tae Kwon Do belt testing. Anyway. Congratulations to Opalg for his correct solution, which follows:

Spoiler

[TIKZ][scale=1.5]
\coordinate [label=left: $A$] (A) at (0,0) ;
\coordinate [label=above right: $B$] (B) at (6,5.04) ;
\coordinate [label=right: $C$] (C) at (6,0) ;
\coordinate [label=above left: $D$] (D) at (40:6) ;
\coordinate [label=right: $E$] (E) at (6,3) ;
\coordinate [label=left: $F$] (F) at (40:4.68) ;
\draw (C) -- (A) -- (B) -- (C) -- (D) -- (E) -- (F) ;
\node at (0.5,0.2) {$\theta$} ;
\node at (4.88,3.5) {$\theta$} ;
\node at (5.825,1) {$\frac12\theta$} ;
\node at (5.825,3.4) {$\frac32\theta$} ;
\node at (3,-0.2) {$1$} ;
\node at (6.2,1.7) {$h$} ;
\node at (5.3,3.6) {$t$} ;
\node at (4.6,2.8) {$l$} ;
[/TIKZ]​

The triangle $ACD$ is isosceles, so $\angle ACD = \frac12\pi - \frac12\theta$. Therefore $\angle BCD = \frac12\theta$, and $\angle BED = \frac32\theta.$

Use a coordinate system with origin at $A$, so that $C = (1,0)$, $B = (1,\tan\theta)$ and $D = (\cos\theta,\sin\theta).$

Let $h = |CE|$, $l = |EF|$ and $t = |DE|.$ Looking at the line $DE$, you see that $E = \left(\cos\theta + t\sin\left(\frac32\theta\right), \sin\theta - t\cos\left(\frac32\theta\right)\right).$ But $E = (1,h).$ Therefore $$\cos\theta + t\sin\left(\tfrac32\theta\right) = 1,\qquad \sin\theta - t\cos\left(\tfrac32\theta\right) = h.$$ Eliminate $t$ from those two equations, getting $\cos\theta\cos\left(\frac32\theta\right) + \sin\theta\sin\left(\frac32\theta\right) = \cos\left(\frac32\theta\right) + h\sin\left(\frac32\theta\right).$ But $\cos\theta\cos\left(\frac32\theta\right) + \sin\theta\sin\left(\frac32\theta\right) = \cos\left(\frac12\theta\right),$ so we get $$h = \frac{\cos\left(\frac12\theta\right) - \cos\left(\frac32\theta\right)}{\sin\left(\frac32\theta\right)}.$$ The triangles $ABC$, $FBE$ are similar, so $$\frac l1 = \frac{\tan\theta - h}{\tan\theta},$$ giving $$l = 1 - \frac h{\tan\theta} = 1 - \frac{\cos\theta\left(\cos\left(\frac12\theta\right) - \cos\left(\frac32\theta\right)\right)} {\sin\theta\sin\left(\frac32\theta\right)}.$$ To evaluate $$\lim_{\theta \searrow 0}\frac{\cos\theta\left(\cos\left(\frac12\theta\right) - \cos\left(\frac32\theta\right)\right)} {\sin\theta\sin\left(\frac32\theta\right)},$$ you can either apply l'Hôpital's rule twice, or (preferably) use the first terms of the power series for sin and cos to get $$\frac{\cos\theta\left(\cos\left(\frac12\theta\right) - \cos\left(\frac32\theta\right)\right)} {\sin\theta\sin\left(\frac32\theta\right)} \approx \frac{\left(1 - \frac{\theta^2}8\right) - \left(1 - \frac{9\theta^2}8\right)}{\frac{3\theta^2}2} = \frac23.$$ Thus $$\lim_{\theta \searrow0}l = 1 - \frac23 = \frac13.$$