# Nested equilateral triangles

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 $$\cos\measuredangle DEC$$...

• @MathLover, edited. I am sorry. Commented Jan 25, 2021 at 19:53

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. $$\cos \alpha = \cos (60^o-\widehat{DGH}) = \cos60^o \cos \widehat{DGH} + \sin60^0 \sin \widehat{DGH} = \frac{4+3\sqrt3}{10}$$

WLOG, $$AB = 8, DF = 5$$. Say $$\angle DEC = \theta$$

If $$CD = x$$ then $$CE = 8 - x$$

Applying sine law in $$\triangle CDE$$,

$$\displaystyle \frac{\sin 60^0}{5} = \frac{\sin \theta}{x} = \frac{\sin(60^0+\theta)}{8-x}$$

From first two, we have $$\sin \theta = \frac{x \sqrt3}{10}$$

From first and third, $$\sin 60^0 \cos \theta + \cos 60^0 \sin\theta = \frac{4\sqrt3}{5} - \frac{x \sqrt3}{10}$$

$$\implies \cos\theta = \frac{8}{5} - \frac{3x}{10}$$

Applying $$sin^2\theta + \cos^2\theta = 1$$, we get

$$x^2 - 8x+\frac{39}{3} = 0 \implies x = 4 \pm \sqrt3$$

So, $$\cos\theta = \frac{4 \mp 3\sqrt3}{10}$$ (please note $$\pm$$ for $$x$$ and corresponding $$\mp$$ for $$\cos \theta$$)

• Thank you for the response! I am not sure I see why $\measuredangle CDE=60^\circ+\theta.$ Actually it isn't? It is equal to $120^\circ-\theta$. Are we using an identity here? What's the point of it? Can't we use $\sin(120^\circ-\theta)$? Commented Mar 6, 2021 at 20:37
• Can you clarify for me how did you get $\sin60^\circ\cos\theta+\cos60^\circ\sin\theta=\dfrac{4\sqrt3}{5}-\dfrac{x\sqrt3}{10}$? Thank you! Commented Mar 6, 2021 at 20:38
• @Katherine from first and third, we have $\displaystyle \frac{\sin 60^0}{5} = \frac{\sin(60^0+\theta)}{8-x}$. Now take $(8-x)$ to LHS and substitute in LHS, $\sin 60^0 = \sqrt3 / 2$. As far as RHS, expand $sin(60^0 + \theta) = \sin 60^0 \cos \theta + \cos 60^0 \sin\theta$. Commented Mar 7, 2021 at 1:59

Note that there are two distinct solutions given the conditions in the problem. This is because there are two admissible orientations of $$\triangle DEF$$: one in which $$AF > BF$$, and one in which $$AF < BF$$. We can understand this by noting that since $$\triangle DEF$$ and $$\triangle ABC$$ share the same center, the circumcircle of $$\triangle DEF$$ will intersect $$\triangle ABC$$ at six points, unless the side length of $$\triangle DEF$$ is less than or equal to half the side length of $$\triangle ABC$$ (in which case there will be either exactly three intersection points in the tangent case, or zero), or greater than or equal to the side length of $$\triangle ABC$$.

A simple way to solve for the required angle is to place the triangle on a coordinate system, say with $$A = (0,0)$$, $$B = (8,0)$$, $$C = (4, 4 \sqrt{3})$$, and let $$F = (x,0)$$ for some $$0 \le x \le 8$$. Then $$E = \left(\frac{8-x}{2}, \frac{(8-x)\sqrt{3}}{2}\right).$$ We then require $$EF = 5$$, or $$25 = \left(x - \frac{8-x}{2}\right)^2 + \left(\frac{(8-x)\sqrt{3}}{2}\right)^2 = 3x^2 - 24x + 64.$$ Consequently $$x = 4 \pm \sqrt{3}$$ and depending on which root is chosen, $$\angle DEC$$ can be determined through a straightforward application of the Law of Sines.

• I was partway through putting together a Geogebra construction of the triangle using the "guess the value of $x$" method. With that in mind I'll just link it here: geogebra.org/calculator/wwf7p4nc. Note that $x=4-\sqrt{3}\approx 2.268$ indeed gives a triangle of the proper size. Commented Jan 25, 2021 at 21:01

You may calculate $$\cos\alpha$$ directly. Per the sine rule for $$\triangle CDE$$ $$\frac{\sin \alpha }{CD}= \frac{\sin (120-\alpha) }{AB-CD}= \frac{\sin 60 }{DE}$$ Eliminate $$CD$$ to get $$\frac{AB}{DE}\sin 60 = \sin\alpha + \sin(120-\alpha)=2\sin 60\cos(60-\alpha)$$ which leads to $$\cos(60-\alpha)=\frac45$$. Then, $$\cos\alpha = \cos(60\pm\cos^{-1}\frac45) =\frac12\cdot \frac45\mp\frac{\sqrt3}2\cdot \frac35 =\frac{4\mp 3\sqrt3}{10}$$

• Thank you for the response! It seems cool, but we still haven't studied how to solve trigonometric equations. Commented Jan 26, 2021 at 9:05
• Can I somehow find $\cos\measuredangle CED$ using what I found $\dfrac{y}{x}=4\pm\sqrt{3}$? Commented Jan 26, 2021 at 9:11
• @nicoledobreva - you can then use the cosine rule to calculate $\cos \angle CED$ fron the ratio $y/x$ Commented Jan 26, 2021 at 16:59
• It is really easier if we assume that $AB=8$ and $DF=5$. We don't need $x$ then. Can I ask you why can we do that? Can we always use this method, e.g. if we have $a:b=c:d$ can we always assume w.l.o.g. that $a=c$ and $b=d$? Commented Jan 26, 2021 at 17:00
• Can I ask you how do we eliminate $CD$? Commented Jan 27, 2021 at 20:40

As others have done, I'll assume $$AB=8,DF=5$$ without loss of generality. Then equilateral triangles $$\triangle ABC$$ and $$\triangle DEF$$ have area $$(\sqrt{3}/4)AF^2 = 16\sqrt{3}$$ and $$(\sqrt{3}/4)DF^2=(25/4)\sqrt{3}$$ respectively. Since the triangles $$\triangle FAE,\triangle DBF, \triangle ECD$$ are all congruent, they must each have area $$\frac{1}{3}\left(16\sqrt{3}-\frac{25}{4}\sqrt{3}\right)=\frac{13\sqrt{3}}{4}.$$ But each triangle has a 60-degree base angle, so we can also write the area of $$\triangle FAE$$ as

$$\frac12 AF\cdot AE \cdot \sin 60^\circ$$

Since $$AE=AC-EC=8-AF$$, this becomes $$\frac12 AF(8-AF)\frac{\sqrt{3}}{2}= \frac{13\sqrt{3}}{4}\implies AF(8-AF)=13$$ with $$x=4\pm \sqrt{3}$$ as solutions. As a minor variation on the other approaches, we may then use the law of cosines on triangle $$\triangle CED$$ to write

$$CD^2=CE^2+ED^2-2CE\cdot ED\cos\measuredangle DEC$$ and therefore

\begin{align} \cos\measuredangle DEC &=\frac{CE^2+ED^2-CD^2}{2CE\cdot ED}\\ &=\frac{(4\pm \sqrt{3})^2+5^2-(4\mp \sqrt{3})^2}{2(5)(4\mp \sqrt{3})}\cdot \frac{4\mp \sqrt{3}}{4\mp \sqrt{3}}\\ &=\frac{(25\pm 16\sqrt{3})(4\mp \sqrt{3})}{130}\\ &=\frac{52\pm 39\sqrt{3}}{130}\\ &=\frac{4\pm 3\sqrt{3}}{10} \end{align} as others have obtained.

• Thank you for the response! I don't understand the first part of your solution (probably you are using some area formulas that I haven't studied). I am new to trig, so I really don't know many things. Can we somehow use what I found - $\dfrac{y}{x}=4\pm\sqrt{3}$ to find $\cos\measuredangle CED$? Commented Jan 26, 2021 at 9:16
• @nicoledobreva To be honest, I looked up the area formulas for equilateral triangles. But these are just applications of the area formula $A=\frac12 ab\sin\theta$ to the case $a=b$ and $\theta=60^\circ.$ As for your work, I’ve effectively assumed $x=1$ and solved for $8-y=4\pm \sqrt{3}$, in agreement with your results. To finish you can combine your results with either law of sines (as others have) or law of cosines (as I have) to get the desired cosine. Commented Jan 26, 2021 at 14:37