Nested Triangles

We note that $\triangle ABC$ , $\triangle BCD$ , and $\triangle CDE$ are similar. Let $BC=a$ , $CA=b$ , and $AB=c$ . From similar triangles,

$\begin{aligned} \frac {AB}{CA} & = \frac {BC}{60} \\ \frac cb & = \frac a{60} \\ \implies c & = \frac {\color{#3D99F6}ab}{60} & \small \color{#3D99F6} \text{Note that }[ABC] = \frac {ab}2 = 3750 \\ & = \frac {7500}{60} = 125 \end{aligned}$

Then we have $\dfrac {[CDE]}{[ABC]} = \dfrac {60^2}{125^2}$ , $\implies [CDE] = \dfrac {60^2}{125^2} \times 3750 = 864$ . Let $EC=x$ and $DE=y$ . Then

$\begin{cases} \dfrac {xy}2 = 864 \implies y = \dfrac {1728}x & ...(1) \\ x^2 + y^2 = 60^2 = 3600 & ...(2) \end{cases}$

From $(2)$ :

$\begin{aligned} x^2 + \color{#3D99F6} \frac {1728^2}{x^2} & = 3600 & \small \color{#3D99F6} \text{Substituting }y = \frac {1728}x \\ x^4 - 3600 x^4 + 2985984 & = 0 & \small \color{#3D99F6} \text{Rearranging and solving} \\ (x^2-1296)(x^2 - 2304) & = 0 & \small \color{#3D99F6} \text{the quadratic for }x^2 \end{aligned}$

$\implies x = \begin{cases} 36 = EC \\ 48 = DE \end{cases}$ . Therefore the perimeter of $\triangle CDE$ is $CD+DE+EC = 60+48+36 = \boxed{144}$ .

@David Vreken This way, we can find out sin(2A) and use it to solve this problem. Pardon me for the slapdash diagram, but you get the idea...

If the area of the entire triangle is $3750$ , then we have that $\frac{CD \times BA}{2}=3750 => \frac{60 \times BA}{2}=3750 => BA=\frac{3750 \times 2}{60}=125$ .

We know that $AC \times BC = CD \times BA = 60 \times 125$ . Rearranging we get $\frac{AC \times BC}{60} = 125$ . However, we also know by Pythagorean theorem that $\sqrt{AC^2+BC^2} = BA = 125$ . At this point, I graphed the two functions ( $\frac{x \times y}{60} = 125$ and $\sqrt{x^2+y^2}=125$ ) and found that the intersect each other at $(75,100)$ . So now we know that $BC=75$ and $AC=100$ .

Next, we can look at the leftmost triangle. To find $BD$ , we can use Pythagorean theorem. $BD=\sqrt{BC^2-DC^2}=\sqrt{75^2-60^2}=45$ . We can subtract this from $BA$ to find $DA$ : $DA=BA-BD=125-45=80$ .

We also know that $CA \times DE = CD \times DA$ . Rearranging to find $DE$ , we get that $DE = \frac{CD \times DA}{CA} = \frac{60 \times 80}{100} = 48$ .

Finally, we can use Pythagorean theorem on the purple triangle: $CD = \sqrt{CD^2-DE^2} = \sqrt{60^2-48^2} = 36$ .

Therefore, the perimeter of $CED$ is $60+48+36=144$ .

From the given informations, we get sin(2A)=0.96. Therefore sin(A)=0.6, cos(A)=0.8 Hence perimeter of the triangle CDE is 60(1+sin(A)+cos(A))=144

I'll start with the observation that $AB = 125$ , given the values of $CD=60$ and the area of triangle $\triangle ABC=3750$ .

Let $x = BD$ . Then $AD = 125-x$ , and given the similarity of triangles $\triangle BDC$ and $\triangle CDA$ , we see that $\frac{60}{x} = \frac{125-x}{60}$ .

Cross-multiplying leads to $3600 = x(125-x)$ , a quadratic equation with solutions $x = 45$ and $x = 80$ . (Aside: the picture suggests that our value of $x$ is the lesser of these two values, but we cannot infer that choice from the information provided. Pictures cannot be used to infer information not explicitly provided. )

If we let $x = 45$ , then we see that all the triangles are similar, with leg proportions in a $3:4:5$ ratio. This leads to (calculations omitted) $BC = 75, AE = 64, CE = 36$ , and $DE = 48$ , with the perimeter of $\triangle CDE = 36 + 48 + 60 = 144$ .

If, however, $x = 80$ , then the proportions on all the triangles are swapped, with the previously shorter legs now the longer legs. This leads to $BC = 100, AE = 27, CE = 48,$ and $DE = 36$ . In this case we also get that the perimeter of $\triangle CDE = 36 + 48 + 60 = 144$ .

Area of triangle ABC = 3750 = (1/2) 60 (BA), so BA = 125. Define BD = x and <DCB = <CDE = <BAC = t. BC = sqrt(3600 + x^2). sin(t) = x/sqrt(3600 + x^2) = sqrt(3600 + x^2)/125. Cross-multiplying, x^2 -125x + 3600 = 0, or (x - 45)(x - 80) = 0, so x = 45 or x = 80. If x = 80, DC = 60, BC = 100, sin(t) = (CE)/60 = 80/100 = .8, and CE = 48,DE = 36, and DE + CE + DC = 144. If x = 45, DC = 60, BC = 75. Sin(t) = 75/125 = .6, = CE /60, and CE = 36. Then DE = 48, and DE + CE + DC = 144.

Let $\angle A = t$ and let $AB = h$ , then $CD = 60 = h \cos t \sin t$ , and $[ABC] = 3750 = \frac{1}{2} h^2 \cos t \sin t$ . Dividing the second equation by the first equation, yields, h = 125, from which $\cos t \sin t = \frac{1}{2} \sin(2t) = 60 / 125$ , or $\sin(2t) = 24/25$ , and this means that $\cos(2t) = \sqrt{25^2 - 24^2}/25 =\frac{ 7}{25}$ . This in turn implies that $\cos t = \sqrt{ (1 + \cos 2 t ) / 2 } = \sqrt{ \frac{1}{2}(1 + \frac{7}{25}) } =\frac{ 4}{5}$ , and therefore, $\sin t = \frac{3}{5}$ . Now $DE = CD \cos t = 60 ( 4/5 )= 48$ , and $CE = CD \sin t = 60 ( 3/5 ) = 36$ , so perimeter $= 60 + 48 + 36 = \boxed{144}$ .