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In ABC right triangle $AC= 2+\sqrt{3}$ and $BC = 3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$

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Here's my strategy of solving this, I'm not sure if it's correct, if you find my explanation hard to understand you can just ignore and write the solution in your own way, thanks.

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1. first area of main triangle, we know AC and CB so it'll be easy to calculate that,

2. to find the radius, we'll reflect triangle ABC on the left side of the circle, turning it into circle inscribed in isosceles triangle, and find it with the formula

3. to find the area of $AMD$ I'll subtract the area of sector $OMD$, triangle $OAD$ and triangle $CDB$ from triangle $AMD$,

4. $DBC$ is an isosceles triangle, so $CB=DB$, then to find the area, I split it into 2 right triangles(it becomes 90 30 60 triangle) and find its height. So we got the Area of $DBC$

5. Now similarly $OAD$ is isosceles, $OD=OC=radius$ of the circle which we "found" also, split this in 2 to get right triangles and then calculate with Pythagorean theorem to find the height so we get Area of $OMD$ too, maybe we could find angles with trigonometry? I don't know that, and if we get the angle of $DOA$ we could find the sector $OMD$ as well and subtract it to the main triangle so we get the area of $AMD$.

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The area $T$ in question can be found as the difference between the area of $\triangle AID$ and a circular segment $IDM$: \begin{align} T&= \tfrac12\,|AI|\cdot|DI|\sin\angle DIA - \frac12\angle DIA\cdot|DI|^2 , \end{align}

where $\angle DIA$ is measured in radians.

To use the standard route to find the radius, consider this circle as inscribed in isosceles $\triangle AFB$, for which

\begin{align} |AC|& = 2+\sqrt3 ,\\ |BC|& = 3+2\sqrt3 ,\\ |AB| &=\sqrt{|AC|^2+|BC|^2} \\ &=\sqrt{28+16\sqrt3}=4+2\sqrt3 ,\\ |FB|&=2|BC|=6+4\sqrt3 ,\\ S_{\triangle AFB}&=|AC|\cdot|BC| =12+7\,\sqrt3 ,\\ \rho&=\tfrac12(2|AB|+|FB|) =|AB|+|BC|=7+4\sqrt3 , \end{align}

\begin{align} r&=|DI|=|MI|=|EI| \\ r &=\frac{S_{\triangle AFB}}{\rho} =\frac{12+7\sqrt3}{7+4\sqrt3} = \frac{(12+7\sqrt3)(7-4\sqrt3)}{(7+4\sqrt3)(7-4\sqrt3)} \\ &=\sqrt3 ,\\ \end{align}

hence, \begin{align} |AI|&=|AC|-|CI|=2 ,\\ \cos\angle DIA&=\frac{|DI|}{|AI|}=\frac{\sqrt3}2 ,\\ \angle DIA&=\frac \pi 6 , \end{align}

and the answer is

\begin{align} T&=\tfrac12\cdot2\cdot\sqrt3\cdot\tfrac12 -\tfrac12\cdot\tfrac{\pi}{6}\cdot(\sqrt3)^2 \\ &=\tfrac14(2\sqrt3-\pi) \approx 0.08062724 . \end{align}

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  • $\begingroup$ nice, but second way of solution gives different answer than first one I think? What I don't understand here is the area of $AID$ triangle, why is the height of the triangle being multiplied by hypotenuse? in your first solution $\endgroup$ Commented Jun 24, 2019 at 20:16
  • $\begingroup$ @bartly ajames: $|AD|=|AI|\cdot\sin\angle DIA$.. $\endgroup$
    – g.kov
    Commented Jun 24, 2019 at 20:25
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Clearly, $\angle CAB = 60$ because $BC = 3+2\sqrt{3} = \sqrt{3} \cdot (2 + \sqrt{3}) = \sqrt{3} \cdot AC$. Now, we need to find the lengths $AM$ and $AD$. Let $r$ be the radius of the circle. It is clear that $OC = r$, and $AO = \sqrt{3}r$ because of the 30-60-90 triangle $AOD$, as $OD$ is a tangent to $AB$. Therefore, $r (1 + \sqrt{3}) = AC = 2 + \sqrt{3} \implies r = \frac{\sqrt{3}+1}{2}$.

Now, we compute the area. Note that $AM = AO - MO = (\sqrt{3}-1)r$, and $AD = \frac{\sqrt{3}}{3}r$, so we have that $[AMD] = AM \cdot AD \cdot \frac{\sqrt{3}}{4} = (\sqrt{3}-1)(\frac{\sqrt{3}}{3})(\frac{\sqrt{3}}{4})r^2 = \frac{1+\sqrt{3}}{8}$.

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  • $\begingroup$ Area of $AMD$ is $\frac{\sqrt3}{2}-\frac{\pi}{4}$ according to test answers, is that correct? $\endgroup$ Commented Jun 24, 2019 at 17:12
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$$\measuredangle B=\arctan\frac{2+\sqrt3}{3+2\sqrt3}=\arctan\frac{2+\sqrt3}{\sqrt3(2+\sqrt3)}=\arctan\frac{1}{\sqrt3}=30^{\circ}.$$

Thus, $$OC=(3+2\sqrt3)\tan15^{\circ}=(3+2\sqrt3)(2-\sqrt3)=\sqrt3(2+\sqrt3)(2-\sqrt3)=\sqrt3$$ and $$\measuredangle AOD=30^{\circ}.$$ Can you end it now?

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  • $\begingroup$ did you calculate that angle for my strategy? so does that mean i could get the answer my way? i didn't really do calculations $\endgroup$ Commented Jun 24, 2019 at 17:13
  • $\begingroup$ @bartly ajames You don't need to get $S_{\Delta BCD}$ because to get $S_{ODBC}$ is much more easier. $\endgroup$ Commented Jun 24, 2019 at 17:37
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Found this solution which looks the simplest to me:

  • $BC = 3 + 2\sqrt{3} = \sqrt{3}AC$;
  • $BD = BC$;
  • $AB = 2AC$;
  • $AD = 1$;
  • $OD = \sqrt{3}$; (tg(<ABC) = AC/BC = AD/OD)
  • $S_{ADO} = \sqrt{3}/2$; (by AD and OD)
  • $S_{MOD} = pi/4;$
  • $S_{AMD} = \sqrt{3}/2 - pi/4;$

p.s. pretty sure there might be better solution by using integrals and (or) function derivatives but that's not mine

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In a video by Mathematics professor Michael Penn, this is solved first by finding out the hypotenuse of the triangle and writing it in the form $a+b\sqrt{3}$, where $a, b \in \mathbb{R}$. Then, he finds the area of the triangle ($AOD$) and the sector ($EOD$) as shown in Figure 1 and subtracts the area of the sector from the triangle to find the shaded area.

Figure 1 Figure 1

However, the shaded area can also be found out using calculus.

We need to find the equation of the line $AB$ and the circle. Before that, we need to find the radius of the circle.

Figure 2 Figure 2

In $\Delta ABC$, let $\angle ABC = \theta$. Then, \begin{align*} \tan \theta &= \frac{AC}{CB}=\frac{1}{\sqrt{3}} \\ \implies \theta &= \frac{\pi}{6} \end{align*} Let us consider $\Delta COB$ and $\Delta DOB$. It can be easily shown that $\Delta COB \cong \Delta DOB$. Hence, $OB$ bisects $\angle ABC$. $$ \therefore \angle CBO = \frac{\pi}{12} $$ In $\Delta CBO$, \begin{align*} \tan{\frac{\pi}{12}} &= \frac{r}{3+2\sqrt{3}} \\ \therefore r &= \sqrt{3} \end{align*} Now, consider a cartesian coordinate system with point $C$ as the origin, as shown in Figure 2. Equation of line $AB$ is given by, \begin{equation} y=f(x)=\frac{(12+7\sqrt{3})-x(2+\sqrt{3})}{(3+2\sqrt{3})} \tag{1}\label{eq:1} \end{equation}

The circle is centred at $(0, \sqrt{3})$. The equation of the circle is, \begin{equation} y=g(x)=\sqrt{3} \pm \sqrt{3-x^2} \tag{2}\label{eq:2} \end{equation} Here, the $\pm$ sign joins the upper half of the circle with the lower half to create the desired circle. But our purpose can be served with the upper half of the circle itself, i.e., the semicircle (The blue semicircular curve as shown in Figure 3. So, we can replace the $\pm$ with $+$ to reduce calculational labour.

Figure 3 Figure 3

If we solve for $x$ and $y$ in equations \eqref{eq:1} and \eqref{eq:2}, we will get the point $D(x, y)$ at which line $AB$ touches the circle (or semicircle).

We get $x=\frac{\sqrt{3}}{2}$ and $y=\frac{3}{2}+\sqrt{3}$. To compute the shaded area, we only need $x$. We will integrate $f(x)$ and $g(x)$ from $x=0$ to $x=\sqrt{3}/2$

Area under the line AB from $x=0$ to $x=\sqrt{3}/2$ is, \begin{align} I_{1} &= \int_{0}^{\sqrt{3}/2} f(x)dx \\ &= \frac{1}{3+2\sqrt{3}}\left(\frac{45\sqrt{3}+78}{8}\right) \\ &= \frac{7\sqrt{3}+12}{8} ~~\text{ sq. unit} \tag{3}\label{eq:3} \end{align}

Also, area under the upper semicircular curve from $x=0$ to $x=\sqrt{3}/2$ is, \begin{align} I_{2} &= \int_{0}^{\sqrt{3}/2} g(x)dx \\ &= \frac{3\sqrt{3}+12}{8} + \frac{\pi}{4} ~~\text{ sq. unit} \tag{4}\label{eq:4} \end{align}

Finally, the required shaded area is given by, \begin{equation} I_{1} - I_{2} = \frac{7\sqrt{3}+12}{8} - \frac{3\sqrt{3}+12}{8} + \frac{\pi}{4} \end{equation} \begin{equation} \boxed{\therefore ~~\text{Reqd. Area} = \left( \frac{\sqrt{3}}{2} - \frac{\pi}{4} \right) ~~\text{ sq. unit}} \tag{5}\label{eq:5} \end{equation}

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