A question about an equilateral triangle

Question

Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.

Answer 1

Let s be the side of the equilateral triangle

Using cosine formula for the triangle

$$\frac{13^2 + 5^2 - s^2}{2.5.13} = \cos\alpha\tag1$$ $$\frac{12^2 + 5^2 - s^2}{2.5.12} = \cos\beta\tag2$$ $$\frac{12^2 + 13^2 - s^2}{2.13.12} = \cos\gamma\tag3$$

Now we have $\gamma + \alpha + \beta = \pi$ $$\cos\gamma = \cos(\pi - \alpha - \beta)$$ Knowing $\cos(\pi - \theta) = \cos(\theta)$

$$\Rightarrow \cos\gamma = \cos(\alpha + \beta)$$ $$\Rightarrow \cos\gamma = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$ $$\Rightarrow (\cos\gamma - \cos\alpha\cos\beta)^2 = \sin^2\alpha\sin^2\beta$$ $$\Rightarrow (\cos\gamma - \cos\alpha\cos\beta)^2 = (1 - \cos^2\alpha)(1 - \cos^2\beta)$$ $$\Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma - 2.\cos\alpha.\cos\beta.\cos\gamma -1 = 0\tag4$$

Applying $(1)$, $(2)$, $(3)$ to $(4)$ and simplifying we have

$$s^4 - 338.s^2 + 17761 = 0$$

using Heron's formula we have

$$\Rightarrow s^2 = \pm 60 . \sqrt3 + 169$$ $$\Rightarrow s = \pm\sqrt{\pm 60 . \sqrt3 + 169}$$

as Area of a triangle is real and positive

$$s = \sqrt{60 . \sqrt3 + 169}$$

Answer 2

Consider an equilateral triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.

Let $P'$, $P'_1$ and $P'_2$ be the reflection of the Point $P$ along $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$

Then we have

$${\triangle APB} \cong {\triangle AP'_2B}$$ $${\triangle APC} \cong {\triangle AP'_1B}$$ $${\triangle BPC} \cong {\triangle BP'C}$$

Thus we can safely say, The area of the hexagon $AP'_1CP'BP'_2 = 2. {\triangle ABC}$

${\triangle AP'_2B}$, ${\triangle AP'_1B}$ and ${\triangle BP'C}$ are isosceles with an apex angle of $120^o$ and composed of $30^o$ and $60^o$ right angle triangle

Thus we have

$$\overline {P'_1P'_2} = 12\sqrt3$$ $$\overline {P'P'_2} = 13\sqrt3$$ $$\overline {P'_1P'} = 5\sqrt3$$

which implies $\angle P'P'_1P'_2 = 90^o$

So the area of the hexagon $AP'_1CP'BP'_2$

$$= \triangle AP'_1P'_2 + \triangle P'P'_1C + \triangle P'P'_2B + \triangle P'P'_1P'_2$$ $$= 12^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + \frac{1}{2}(5\sqrt3)(12\sqrt3)$$ $$=169\frac{\sqrt3}{2} + 90$$ $${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$

Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then

$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$

$$\frac{\sqrt3}{4}a^2 = 169\frac{\sqrt3}{4} + 45$$ $$a^2 = 169 + 45\frac{4}{\sqrt3}$$ $$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$

Simplifying

$$a = \sqrt{169 + 60\sqrt3}$$

Answer 3

Consider an equilateral triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.

Rotate $P$ clockwise along $A$ by $60^o$ to $B'$, along $B$ by $60^o$ to $C'$, along $C$ by $60^o$ to $A'$

We note that, $${\triangle APB} \cong ${\triangle ACB'}$$ $${\triangle APC} \cong ${\triangle A'CB}$$ $${\triangle BPC} \cong ${\triangle AC'B}$$

Thus we have, the area of the hexagon $AB'CA'BC'$ $$=2.{\triangle ABC}$$

We also observe that ${\triangle APB'}$, ${\triangle BPC'}$ and ${\triangle A'PC}$ are equilateral triangles

${\triangle PCB'} \cong {\triangle AC'P} \cong {\triangle A'BP} = RightAngle {\triangle}$

Thus we can say

$$2.{\triangle ABC} = [AB'CA'BC']$$ $$= {\triangle APB'} + {\triangle BPC'} + {\triangle A'PC} + 3.{\triangle PCB'}$$ $$= 12^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4} + 3.\frac{1}{2}(5)(12)$$ $$=169\frac{\sqrt3}{2} + 90$$ $${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$

Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then

$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$

$$\frac{\sqrt3}{4}a^2 = 169\frac{\sqrt3}{4} + 45$$ $$a^2 = 169 + 45\frac{4}{\sqrt3}$$ $$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$

Simplifying

$$a = \sqrt{169 + 60\sqrt3}$$

Answer 4

The Wikipedia article on equilateral triangles quotes the following theorem from "Curious Properties of the Circumcircle and Incircle of an Equilateral Triangle," by Prithwijit De (http://ms.appliedprobability.org/data/files/Abstracts%2041/41-1-7.pdf):

Theorem: Let $ABC$ be an equilateral triangle with side $s$, and let $P$ be a point in the plane of the triangle with distances $p$, $q$, and $r$ to $A$, $B$, and $C$, respectively. Then $$3(p^4+q^4+r^4+s^4) = (p^2+q^2+r^2+s^2)^2{\textrm.}$$ Solving $3(5^4+12^4+13^4+s^4) = (5^2+12^2+13^2+s^2)^2$ for $s$ gives $s=\sqrt{169\pm 60\sqrt3}$. Because $P$ is inside the triangle, $s>13$, which allows only $s=\sqrt{169+ 60\sqrt3}$, the same solution others have obtained.

Answer 5

If we are allowed to use Coordinate Geometry,

WLOG we can assume the coordinates of $A,B,C$ to be $(x,y),(-a,0),(a,0)$ respectively and $D$ to be $(h,k).$

As $\triangle ABC$ is equilateral $(x-a)^2+(y-0)^2=(x+a)^2+(y-0)^2=(-a-a)^2+(0-0)^2$

From the first relation we find $4ax=0\implies x=0$ as $a\ne0$

From the second relation we find $y^2=3a^2\implies y=\sqrt3a$ assuming $a>0$

So, $\overline{AB}=2a$

So, $\overline{AD}^2=(h-0)^2+(k-\sqrt3a)^2$ But $\overline{AD}=13$ $\implies (h-0)^2+(k-\sqrt3a)^2=13^2--->(1)$

Similarly, for $BD,CD$ respectively,

$(h+a)^2+k^2=12^2--->(2)$ and $(h-a)^2+k^2=5^2--->(3)$

So here we have three equations with three unknowns $a,h,k$

One way to solve this could be as follows :

From $(2)-(3),4ah=119\implies h=\frac{119}{4a}$

From $(2)-(1),2ah+2\sqrt3ak=2a^2-25\implies k=\frac{4a^2-169}{4\sqrt3a} $ (putting $ah=\frac{119}4$)

Putting the values of $h,k$ in terms of $a$ in $(1),$

$$\left(\frac{119}{4a}\right)^2+\left(\frac{4a^2-169}{4\sqrt3a}-\sqrt3a\right)^2=13^2$$

$$\implies 64a^4-32\cdot169a^2+169^2+3(119^2)=0$$

which is a Quadratic Equation in $a^2$