Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.
Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.
Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$…
Best Answer
Here I will show how you can draw the figure. The calculations that follow are somewhat similar to other answers.
We can build triangle $DEF$ by scaling down triangle $ABC$ and rotating it around its centroid $G$ by some angle $\alpha$.
Notice the hatched triangle $DHG$. This is a right triangle with sides: $$GH = \frac13 AH =\frac{4}{12} AH$$ $$DG = \frac23 DY = \frac23 \times \frac58 AH = \frac{5}{12} AH $$ $$HD = \sqrt{DG^2 - GH^2} = \frac{3}{12} AH$$ It seems the designer of the problem intentionally used the ratio $\frac58$ to lead us to the $3-4-5$ right triangle!
Anyway, Now we are able to locate point $D$ on $BC$. It is at distance $\frac14 AH$ from point $H$ , the midpoint of BC. We can similarly locate points $E$ and $F$ , hence constructing triangle $DEF$.
Now, as for calculations, note that $\widehat{DEC}=120^o - \alpha$ and that $\widehat{DGH}=60^o - \alpha$. So $\widehat{DEC}=60^o+\widehat{DGH}$ . Therefore: $$\cos \widehat{DEC} = \cos (60^o+\widehat{DGH}) = \cos60^o \cos \widehat{DGH} - \sin60^0 \sin \widehat{DGH}$$ $$= \frac12 \times \frac45 - \frac{\sqrt 3}{2} \times \frac35 = \frac{4-3\sqrt3}{10}$$ Note that when locating point $D$ we have two options, say, above and below $H$. The above calculations are for the option shown in the figure, where $D$ is selected below $H$. Alternatively, we could select $D$ to be above $H$, and respectively have $E$ below $K$. The interested reader can verify that in that case $\widehat{DEC}$ would be $\alpha$. And $\; \cos\alpha \;$ can be calculated similar to above. $$\cos \alpha = \cos (60^o-\widehat{DGH}) = \cos60^o \cos \widehat{DGH} + \sin60^0 \sin \widehat{DGH} = \frac{4+3\sqrt3}{10}$$