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January 22nd, 2018, 06:05 AM   #1
Joined: Jan 2018
From: Belgrade

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Problem with square, circle and triangle

I have the following problem:

In the area of the square $\displaystyle ABCD$, a semicircle with center $\displaystyle O$ and radius $\displaystyle r$ is constructed over the $\displaystyle AB=a$ as the diameter. Tangent $\displaystyle CT$ touches semicircle at the point $\displaystyle T$. If half-line $\displaystyle OT$ intersects $\displaystyle AD$ in point $\displaystyle T$, what is the area of the triangle $\displaystyle OCE$?
I solved the problem this way:

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?
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January 22nd, 2018, 02:55 PM   #2
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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 $\displaystyle \frac{5}{12}a^2$.
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January 22nd, 2018, 11:06 PM   #3
Joined: Jan 2018
From: Belgrade

Posts: 55
Thanks: 2

You wrote OT intersects AD in point T, but it's clear that what you intended was OT extended meets AD in point E.
Yes, that's it. I thought it was clear enough by saying $\displaystyle OT$ is half-line (ray).

Your b should have been 2.
That's right. Thanks. My mistake.

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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