Let $ABC$ be an equilateral triangle with side length $2$. Let the circle with diameter AB be $\gamma$. Consider the two tangents from $C$ to $\gamma$, and let the tangency point closer to $A$ be $D$. Find the area of triangle $CAD$.

I was able to figure out that $CD$ has to be $\sqrt2$. I can not figure out the height of triangle $CAD$.

I am trying to calculate the height from $D$ to $AC.$ Alternatively, if I could find the length of $AD$, then I should also be able to solve problem.

  • $\begingroup$ but you know all three angles are $60$ degrees. if you divide one side in half and then work from there. but maybe you dont even need to divide. $\endgroup$ – Natural Number Guy Mar 12 '19 at 3:33
  • $\begingroup$ Can you use trigonometry? If you know $AC$, $CD$ and angle between them, you should be able to find the area. Although I think $CD=1$. $\endgroup$ – Vasya Mar 12 '19 at 3:39
  • $\begingroup$ so, since all sides are of equal size, and from pytagorean theorem: $a^2+b^2=c^2$. then $h^2 = a^2 - (\frac{a}{2})^2$. where $h$ is the height. that will give you a start. but there are more than one way to do find area though. $\endgroup$ – Natural Number Guy Mar 12 '19 at 3:39
  • $\begingroup$ bah, I worked on a solution and read wrong (area of $ABC$ instead of $CAD$). $\endgroup$ – Natural Number Guy Mar 12 '19 at 18:06

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\begin{align} S_{ACD}&= S_{CDO}+S_{AOD}-S_{AOC} ,\\ S_{CDO}&=\tfrac12|CD|\cdot|OD|=\frac{\sqrt2}2 ,\\ S_{AOD}&=\tfrac12|OA|\cdot|OD|\cdot\sin\angle AOD =\tfrac12\sin\angle OCD =\tfrac12\cdot\frac{|OD|}{|OC|} =\frac{\sqrt3}6 ,\\ S_{AOC}&=\tfrac12|OA|\cdot|OC|=\frac{\sqrt3}2 ,\\ S_{ACD}&=\frac{\sqrt2}2+\frac{\sqrt3}6- \frac{\sqrt3}2 =\frac{\sqrt2}2-\frac{\sqrt3}3 \approx .1297565117 . \end{align}

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Let $O$ be a center of the circle $\Gamma$.

Thus, by the Pythagoras's theorem $$CD^2=CO^2-DO^2=\left(\sqrt3\right)^2-1^2=2$$ and $$CD=\sqrt2.$$ Id est, $$S_{\Delta ACD}=\frac{1}{2}CD\cdot AC\sin\measuredangle ACD=$$ $$=\frac{1}{2}\cdot\sqrt2\cdot2\sin\left(\arctan\frac{1}{\sqrt2}-30^{\circ}\right)=$$ $$=\sqrt2\left(\sin\arctan\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}-\cos\arctan\frac{1}{\sqrt2}\cdot\frac{1}{2}\right)=$$ $$=\sqrt2\left(\frac{\frac{1}{\sqrt2}}{\sqrt{1+\left(\frac{1}{\sqrt2}\right)^2}}\cdot\frac{\sqrt3}{2}-\frac{1}{\sqrt{1+\left(\frac{1}{\sqrt2}\right)^2}}\cdot\frac{1}{2}\right)=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$

There is a solution without trigonometry.

Let $DK$ be an altitude of $\Delta ADO$.

Thus, since $\Delta OKD\sim\Delta CDO,$ we obtain: $$\frac{DK}{DO}=\frac{DO}{CO}$$ or $$\frac{DK}{1}=\frac{1}{\sqrt3},$$ which gives $$DK=\frac{1}{\sqrt3}$$ and $$S_{\Delta ACD}=S_{\Delta ADO}+S_{\Delta OCD}-S_{\Delta ACO}=$$ $$=\frac{1}{2}\cdot\frac{1}{\sqrt3}\cdot1+\frac{1}{2}\cdot\sqrt2\cdot1-\frac{1}{2}\cdot\sqrt3\cdot1=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$

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  • $\begingroup$ Thanks for above. I should have tried to use trigonometry. My student only knows geometry and the above problem was under geometry section, so I was trying to solve with just geometry. Above makes much more sense than trying to solve with just geometry. $\endgroup$ – user653261 Mar 12 '19 at 4:55
  • $\begingroup$ @user653261 I added the geometric solution. $\endgroup$ – Michael Rozenberg Mar 12 '19 at 6:54

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