Three Homogeneous Quadratics

Note that the cosine rule is: $a^2=b^2+c^2-2bc\text{cos} \angle A$ If $\angle A=120^\circ$ , we will have $\text{cos} \angle A = -\frac{1}{2}$ In this special case, we have $a^2=b^2+c^2-2bc(-\frac{1}{2})=b^2+c^2+bc$ .

Consider the cartesian plan, with $O$ as the origin. Points $A, B, C$ are such that $OA, OB, OC$ make $120^\circ$ between each other, and we have lengths $OA= j, OB=k, OC=l$ . Referring back to the question, $j^2+k^2+jk=9$ refers to a triangle with sides $j, k$ , angle between them $120^\circ$ , with a third side of 3. Hence we have $AB=3$ . Similarily, equations $k^2+l^2+kl=16$ and $l^2+j^2+lj=25$ gives us $BC = 4, CA = 5$ . The interior point of the triangle defined as the point where we place the three $120^\circ$ angles together will be called P.

By definition, P is the Fermat point of ABC. Note $\triangle ABC$ is a right triangle, by the Pythagorean Theorem. To compute $j+k+l$ , place point Q so that BCQ is equilateral with side 4, $\angle ABQ$ is obtuse. We now will show that $j+k+l=CQ$ . Let R be the point on CQ such that $\angle BRC=120^\circ$ . Note that QBRC is a cyclic quadrilateral, since $\angle BRC+\angle BQC=180^\circ$ . Thus $\angle QRB=\angle QCB=60^\circ$ . Let S be on QR such that $RS=RB$ . Note that since $\angle QRB=\angle SRB=60^\circ$ , we see $\triangle SRB$ is an equilateral triangle. By QBRC as a cyclic quadrilateral, since $\angle RQB=\angle RCB$ . Since $QB=CB$ , $SB=RB$ , $\triangle QSB$ is conguruent to $\triangle CRB$ . Thus $QS=CR$ . By $BR=SR$ , we see $CR+BR+AR=QA$ . Since $\angle CRB=\angle CRA=\angle ARB=120^\circ$ by chasing angles, we note that R is defined at the same place as P. Thus $j+k+l=QA$ By cosine rule, $(j+k+l)^2=(QA)^2=3^2+4^2+2 \times 3 \times 4 \times \text{cos}150^\circ=25+12\sqrt{3}$ Thus $a+b+c=25+12+3=40$

[Edits for clarity - Calvin]

Consider the fact that $\cos(120^\circ)=-0.5$ . So, $k^2+j^2+kj=k^2+j^2- 2(k)(j)\cos(120^\circ)=9=3^2$ Similarly we look at $j^2+l^2-2(j)(l)\cos(120^\circ)=5^2, k^2+l^2-2(k)(l)\cos(120^\circ)=4^2.$ Now consider triangle ABC with an internal point O such that $AO=j, BO=k, OC=l$ and angle AOB=angle BOC=angle COA =120 degrees.

By applying cosine rule to triangle ABC, and with the equations from above, we will get that AB=3 and BC=4 and AC=5. We note that triangle ABC is a right angle triangle at which it has an area of $1/2 * 3 * 4 =6$ Also, Area of triangle ABC = $(1/2)\sin(120^\circ) * (kj+kl+jl)=6$ So we get $kj+kl+jl=8\sqrt{3}$ .

Adding the three original equations together we will get; $2(k^2+j^2+l^2)+kj+kl+jl=9+16+25=50$ . Now adding 3 times of $kj+kl+jl=8\sqrt{3}$ to it, we will get; $2(k^2+j^2+l^2)+4(kj+kl+jl)=50+24\sqrt{3}$ . Since $2(k^2+j^2+l^2)+4(kj+kl+jl)=2(k+j+l)^2$ , so $2(k+j+l)^2=50+24\sqrt{3}$ or $(k+j+l)^2=25+12\sqrt{3}$ Hence our required answer is $25+12+3=40$ .

[Minor Latex edits - Calvin]

From the three equations we known, we can have (l-j)(j+l+k)=7,(j-k)(j+l+k)=9,(l-k)(j+l+k)=16 We can suppose (j+l+k)=x, then we have l-j=7/x, j-k=9/x. So j=k+9/x, l=k+16/x. Then I+j+k=3k+25/x, so we have 3kx=x^2-25. Because j^2+k^2+jk=9 and j=k+9/x so we have k^2+18k/x+81/x^2+k^2+k^2+9k/x=9, same to 3k^2+27k/x+81/x^2=9 It is also same to 3k^2x^2+27kx+81=9x^2 and 9k^2x^2+81kx+243=27x^2, bring 3kx=x^2-25 in it we can have (x^2-25)^2+27(x^2-25)+243=27x^2 x^4-50x^2+625-27*25+243=0 x^4-50x^2+193=0 From the Root Formula x^2=25 \pm 12 \sqrt{3} So a+b+c =25+12+3=40

Noticed that

$j^2+k^2+jk=j^2+k^2-2jk\cos{120^\circ}=3^2$

$k^2+l^2+kl=k^2+l^2-2kl\cos{120^\circ}=4^2$

$l^2+j^2+lj=l^2+j^2-2lj\cos{120^\circ}=5^2$

With these, we can construct a right triangle

HomogeneousQuadratic

with $\angle ADB=\angle BDC=\angle CDA=120^\circ$ .

From the area of triangle, we can get

$\frac{1}{2}jk\sin120^\circ+\frac{1}{2}kl\sin120^\circ+\frac{1}{2}lj\sin120^\circ=6$

$\frac{\sqrt3}{4}(jk+kl+lj)=6$

$\boxed{jk+kl+lj=8\sqrt3}$

Sum up the three given equations, we have

$2(j^2+k^2+l^2)+(jk+kl+lj)=50$

Substitute $jk+kl+lj=8\sqrt3$ to the equation we have

$\boxed{j^2+k^2+l^2=25-4\sqrt3}$

Now, we have enough informations to solve

$(j+k+l)^2$

$=(j^2+k^2+l^2)+2(jk+kl+lj)$

$=(25-4\sqrt3)+2(8\sqrt3)$

$=25+12\sqrt3$

The idea to this problem is not simple enough. Firstly I tried to solve this problem using AM -GM inequality trying to show that $(j+k+l)^2 \geq k$ and $(j+k+l)^2 \leq k$ for some $k$ , implying that $(j+k+l)^2 = k$ . But that did not work out. The idea to this problem is transforming the problem to a geometric problem. Carefully look at $j^2 + jk + k^2$ .

Dosen't it look like that in a triangle, 2 sides are of length $j$ units and $k$ units with an included angle of $120^{\circ}$ and we are trying to compute the third side? Yes it indeed does. We now use the idea of Fermat Point.

A little observation reveals that $j$ , $k$ and $l$ are the distances of the Fermat point from the three vertices of a triangle.Thus we have grabbed the main idea of the problem and we proceed as follows.

Take a point $P$ in a plane and construct line segments of length $j,k,l$ pairwise having an included angle of $120^{\circ}$
Mark the extremities of the segments as $A$ , $B$ ans $C$ . [ Since we dont know $j,k,l$ , the construction will not be accurate] . Thus

$AB^2 = j^2 + k^2 - 2 j k \cos120^{\circ} = j^2 + k^2 + jk = 9$ . Hence, $AB = 3$ . Similarly , $BC = 4$ and $AC = 5$ .

From converse of Pythagorus's Theorem, we find that $\angle ABC = 90^{\circ}$ .

So, $P$ is the Fermat Point of a triangle with side lengths $3,4,5$ . Now we need to compute $PA + PB + PC$

We construct an equilateral triangle with $BC$ as one of is sides outwardly. Let the triangle be $BPA_1$ . Similarly construct equilateral triangles $ACB_1$ and $ABC_1$ . It is a well known result that $AA_1, BB_1 , CC_1$ are concurrent at the Fermat point of the triangle. To get exact constructions, first construct the triangle with sides $3,4,5$ and then construct the equilateral triangles outwardly. Then trace the Fermat point. But we ned not do tha as its just an algebra problem and we dont need to be accurate. Moreover, $AA_1= BB_1 = CC_1 = PA + PB + PC$ ( One may google it if he/she dosent know) ( http://www.math.washington.edu/~king/coursedir/m444a03/notes/12-05-Napoleon-Fermat.html) and http://www.cut-the-knot.org/Generalization/fermat_point.shtml) might be helpful. Image Thus it is enough to compute $AA_1^2$ as it is essentially equal to $(PA+Pb+PC)^2 = (j+k+l)^2$

Now look at $\triangle ABA_1$ . We have $AB = 3$ , $BA_1 = BC = 4$ and $\angle ABA_1 = (90+60)^{\circ} = 150^{\circ}$

We have, by applying Cosine rule to $\triangle ABB_1$

$AA_1^2 = AB^2 + BA_1^2 - 2 \times AB \times BA_1\times \cos \angle ABA_1$

$\implies$ $AA_1^2 = 3^2 + 4^2 - 2 \times 3 \times 4 \times \cos 150^{\circ}$

$\implies$ $AA_1^2 = 9 + 16 + 2 \times 3 \times 4 \times \cos 30^{\circ}$ (as $\cos \theta = - \cos ( 180^{\circ} - \theta)$ )

$\implies$ $AA_1^2 = 25 + 2 \times 3 \times 4 \times \frac{\sqrt{3}}{2} = 25 + 12\sqrt{3}$

$\implies$ $(j+k+l)^2 = 25 + 12\sqrt{3}$ and hence the answer is $\boxed{25+12+3=40}$

Note- There may be non geometric solutions, but i could not come up with anything else.

First of all, do note that by adding up the three equations, we get 2(j^2+k^2+l^2) + jk+kl+jl = 50. Hence jk+kl+jl = 50 - 2(j^2+k^2+l^2). -------(1)

Now by direct expansion, (j+k+l)^2 = j^2+k^2+l^2 + 2(jk+kl+jl) ----(2) Substituting (1) into (2), we have (j+k+l)^2 = 100 - 3(j^2+k^2+l^2) ----(3)

Now let's imagine all three equations as quadratic equations: In j^2 + jk +(k^2-9) = 0, let j be the variable and k, the constant: By the quadratic formula, we have j = (\sqrt{36-3k^2} -k)/2 Reversing roles (i.e. k is the variable and j is the constant), k= (\sqrt{36-3j^2} -j)/2.

Applying the same rules for k^2 + kl +(l^2-16) = 0, we have k= (\sqrt{64-3l^2} -l)/2 and l = (\sqrt{64-3k^2} -k)/2.

Similarly, in l^2 + j^2 +jl =25, we have l= (\sqrt{100-3j^2} -j)/2 and j = (\sqrt{100-3l^2} -l)/2.

Now expressing equation (3) in terms of j by direct substitution and manipulation, we will get (j+k+l)^2 = 3j[(\sqrt{100-3j^2})+(\sqrt{36-3j^2} -j)]/2. -------(4) Now multiplying the first and last original equations by 4, we have: 4j^2+4k^2+4jk = 36 and 4l^2+4j^2 + 4jl = 100 which reduces equation (4) to (j+k+l)^2 =3j(j+k+l)-2. Making j the subject, j= (j+k+l)/3 +2/[3(j+k+l)]

Using the same method for k, one would find that k= (j+k+l)/3 -25/[3(j+k+l)]. Substituting these two values into j^2+k^2 +jk = 9, we have a quartic equation: (j+k+l)^4 - 50(j+k+l)^2 +193 = 0. Let (j+k+l)^2 = x. Then x^2- 50x +193 = 0 By the quadratic formula, the value of x can be reduced to 25+12\sqrt{3} where a=25, b=12, c=3. Hence a+b+c = 40.