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January 22nd, 2018, 07:05 AM  #1  
Newbie Joined: Jan 2018 From: Belgrade Posts: 4 Thanks: 0  Problem with square, circle and triangle
I have the following problem: Quote:
We see that $\displaystyle OB=\frac{a}{b}$, so from $\displaystyle \triangle OBC$: $\displaystyle OC=\frac{a}{2}\sqrt{5}$ and from $\displaystyle \triangle OCT$: $\displaystyle CT=a \: (OT=\frac{a}{2})$. Let us denote $\displaystyle \angle BCO = \frac{\beta }{2}$. $\displaystyle \triangle OBC$ and $\displaystyle \triangle OCT$ are congruent, so $\displaystyle \angle OCT = \angle BCO = \frac{\beta }{2}$ and $\displaystyle \angle BCT = \beta$. $\displaystyle \angle BCT$ and $\displaystyle \angle AOE$ are sharp angles with normal rays, so $\displaystyle \angle AOE = \angle BCT = \beta$. Using trigonometrical identity $\displaystyle \cos \beta = 2\cos^{2} \frac{\beta }{2}  1$ and $\displaystyle OE = \frac{OA}{\cos \beta } = \frac{\frac{a}{2}}{\cos \beta }$, we get $\displaystyle OE = \frac{5}{6}a$ and area of $\displaystyle \triangle OCE$: $\displaystyle P_{\triangle OCE} = \frac{1}{2}OE \cdot CT = \frac{5}{12}a^{2}$ I'd like to solve this problem in an easier way, without using trigonometry. Is it possible?  
January 22nd, 2018, 03:55 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,557 Thanks: 1480 
You wrote $OT$ intersects $AD$ in point $T$, but it's clear that what you intended was $OT$ extended meets $AD$ in point $E$. Your solution is essentially correct. Your $b$ should have been $2$. It's possible to solve the problem without using trigonometry, but I'll leave it to you to judge whether the method below is easier than yours. Let $CT$ extended meet $AE$ in the point $F$. As triangles $OTF$, $CBO$ are similar and $OT/CB = 1/2$, $FA = FT = a/4$. As triangles $FTE$, $OAE$ are similar, $TE/FT = AE/OA$. Hence $TE/(a/4) = AE/(a/2)$, which implies $AE = 2TE$, and so $FE = 2TE  a/4$. By Pythagoras for triangle $EFT$, $TE^2 + (a/4)^2 = (2TE  a/4)^2$. Hence $TE = a/3$, and so $OE = a/2 + a/3 = 5a/6$. Hence triangle $OCE$ has area $\displaystyle \frac{5}{12}a^2$. Semicircle.JPG 
January 23rd, 2018, 12:06 AM  #3  
Newbie Joined: Jan 2018 From: Belgrade Posts: 4 Thanks: 0  Quote:
Quote:
Can you explaine in more detail that triangles $\displaystyle OTF$ and $\displaystyle CBO$ are similar. I can't see that, but everything else is clear. I think solution that uses trigonometry is less complicated. But, this problem is for those enrolling high school. In that age they don't learn trigonometry yet. That's why I asked for solution that suites them better.  

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