# Geometry high school math competition question

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.

• 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. – Natural Number Guy Mar 12 at 3:33
• Can you use trigonometry? If you know $AC$, $CD$ and angle between them, you should be able to find the area. Although I think $CD=1$. – Vasya Mar 12 at 3:39
• 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. – Natural Number Guy Mar 12 at 3:39
• bah, I worked on a solution and read wrong (area of $ABC$ instead of $CAD$). – Natural Number Guy Mar 12 at 18:06

\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}

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 $$S_{\Delta ACD}=S_{\Delta ADO}+S_{\Delta OCD}-S_{\Delta ACO}=$$ $$=\frac{1}{2}\cdot\frac{1}{\sqrt3}\cdot1+\frac{1}{2}\cdot\sqrt2\cdot1-\frac{1}{2}\cdot\sqrt3\cdot1=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$

• 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. – user653261 Mar 12 at 4:55
• @user653261 I added the geometric solution. – Michael Rozenberg Mar 12 at 6:54