An algebra problem by Sourabh Jangid

Looking at the terms in the system of equations, $\sqrt{ y^2 - \frac{1}{16} }$ remind us of calculating the length of the opposite of a right angled triangle with adjacent side $\frac{1}{4}$ and hypotenuse $y$ . Combined with the other term $\sqrt{ z^2 - \frac{1}{16} }$ , this induces us to think placing 2 right triangle with a common base of $\frac{1}{4}$ together as in the figure on the right. Then, the equation states the the sum of the opposite sides will be $x$ . Notice that in this triangle, the altitude is $\frac{1}{4}$ .

Using this as inspiration, let's consider a triangle with altitudes $\frac{1}{4}, \frac{1}{5}$ and $\frac{1}{6}$ . For now, let's assume that 1) the triangle exists and 2) the triangle is acute. Proofs of these assumptions are given at the end.

Claim: The side lengths of this triangle will satisfy the given system of equations.

Proof: Let the perpendicular from $A$ meet $BC$ at $D$ . Since the triangle is acute, hence $D$ lies between $B$ and $C$ . As such,

$a = BC = BD + DC = \sqrt{ BA^2 - AD^2 } + \sqrt{ CA^2 - AD^2 } = \sqrt{ c^2 - \frac{1}{16} } + \sqrt{ b^2 - \frac{1}{16} }$

Similarly, the other sides yield the corresponding equations. $_\square$

Now, let's determine the sides of the triangle. By considering it's area both in terms of 1/2 (base) (height) and heron's formula , we get that
$\text{ Area} = \frac{x}{8} = \frac{y}{10} = \frac{z}{12} = \frac{1}{4} \sqrt{ ( x+y+z)(x+y-z)(x-y+z)(-x+y+z) }$

Substituting $y = \frac{5}{4} x, z = \frac{6}{4} x$ , we get that

$\frac{x}{8} = \frac{1}{4} \sqrt{ x^4 ( 1 + \frac{5}{4} + \frac{6}{4} ) ( 1 + \frac{5}{4} - \frac{6}{4} ) ( 1 - \frac{5}{4} + \frac{6}{4} ) ( - 1 + \frac{5}{4} + \frac{6}{4} ) } = \frac{15\sqrt{7}}{64} x^2$

Since $x \neq 0$ , thus $x = \frac{ 8 } { 15 \sqrt{ 7 }}$ and so $x + y + z = \frac{15}{4} x = \frac{ 2}{ \sqrt{7} }$ .

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We now prove the 2 assumptions

Assumption 1: Such a triangle exists.

Proof: To show that the triangle exists, we simply need to show that the sides satisfy the triangle inequality. Let the sides corresponding to the altitude of $\frac{1}{4}, \frac{1}{5}$ and $\frac{1}{36}$ be $a, b$ and $c$ . We know that "side * altitude = area / 2", hence we conclude that $a : b : c = 5 \times 6 : 4 \times 6 : 4 \times 5$ . Now, since $24 + 20 > 30$ , hence the values satisfy the triangle inequality. $_\square$

Assumption 2: Such a triangle is acute.

Proof: Similarly, from cosine rule , we need to show that $a^2 + b^2 > c^2$ . This is true since $24^2 + 20^2 > 30 ^2$ . $_\square$

Let $x_1 = \sqrt{y^2 - \frac{1}{16}}, x_2 = \sqrt{z^2 - \frac{1}{16}}, y_1 = \sqrt{z^2 - \frac{1}{25}}, y_2 = \sqrt{x^2 - \frac{1}{25}}, z_1 = \sqrt{x^2 - \frac{1}{36}}, z_2 = \sqrt{y^2 - \frac{1}{36}}$ . Thus $x = x_1 + x_2, y = y_1 + y_2, z = z_1 + z_2$ , all of which are positive real numbers. Draw a line segment $\overline{AB}$ of length $x$ , and locate $D$ on $\overline{AB}$ such that $AD = x_1$ and $DB = x_2$ . Construct a perpendicular $\overline{CD}$ to $\overline{AB}$ of length $\frac{1}{4}$ , then draw line segments $\overline{AC}$ and $\overline{BC}$ .

By the above equations $(AC)^2 = x_1^2 + \frac{1}{16} = y^2 \implies AC = y$ , and $(BC)^2 = x_2^2 + \frac{1}{16} = z^2 \implies BC = z$ . Therefore $\Delta ABC$ has sides of length $x$ , $y$ , and $z$ . Furthermore, $\angle ABC$ and $\angle BAC$ are acute. By a symmetrical construction, we can create triangles of base $y$ and altitude $\frac{1}{5}$ , and of base $z$ and altitude $\frac{1}{6}$ , whose sides are of length $x$ , $y$ , and $z$ , which by $SSS$ are congruent. In each case the angles adjacent to the base are acute, so $\Delta ABC$ is acute with altitudes $\frac{1}{4}$ , $\frac{1}{5}$ , and $\frac{1}{6}$ .

To determine the values of $x$ , $y$ , and $z$ , we may proceed as in Calvin Lin's proof by using Heron's formula to find the area of $\Delta ABC$ in terms of $x$ , etc., but here is an alternative method. There is a formula that is similar to Heron's which gives the area of a triangle in terms of its altitudes instead of its sides. Let $A$ be the area of the triangle, $h_a, h_b, h_c$ the altitudes, and $H = \frac{1}{2}(h_a^{-1} + h_b^{-1} + h_c^{-1})$ . Then

$A^{-1} = 4\sqrt{H(H - h_a^{-1})(H - h_b^{-1})(H - h_c^{-1})}$ .

For our triangle, let $h_a = \frac{1}{4}$ , $h_b = \frac{1}{5}$ , $h_c = \frac{1}{6}$ . Thus $H = \frac{1}{2}(4 + 5 + 6) = \frac{15}{2}$ and

$A^{-1} = 4\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}} = 15\sqrt{7} \implies A = \frac{1}{15\sqrt{7}} = \frac{\sqrt{7}}{105}$ .

Finally, computing the area of the triangle via the formula $A = \frac{1}{2}bh$ using each side in turn as the base gives us $x$ , $y$ , and $z$ :

$\frac{\sqrt{7}}{105} = \frac{1}{2}x\frac{1}{4} = \frac{x}{8} \implies x = \frac{8\sqrt{7}}{105}$

$\frac{\sqrt{7}}{105} = \frac{1}{2}y\frac{1}{5} = \frac{y}{10} \implies y = \frac{10\sqrt{7}}{105}$

$\frac{\sqrt{7}}{105} = \frac{1}{2}z\frac{1}{6} = \frac{x}{12} \implies z = \frac{12\sqrt{7}}{105}$

$\therefore x + y + z = \frac{30\sqrt{7}}{105} = \frac{2\sqrt{7}}{7} = \frac{2}{\sqrt{7}} \quad\square$

In each square root, we have a difference of 2 squares. This would motivate us to find a geometric interpretation of the problem, using Pythagoras' Theorem to help us.

Image taken from GeoGebra .

Indeed, consider an acute triangle with sides $x,y,z$ . If we let $H_k$ denote the length of the altitude from a vertex to the side with length $k$ , what can we say about its altitudes $H_x,H_y,H_z$ ? From the equations, we can tell that $H_x=\frac14,H_y=\frac15, H_z=\frac16$

$\mbox{Area}=\frac12 \left(\frac14 \right)(x)=\frac12 \left(\frac15 \right)(y)=\frac12 \left(\frac16 \right)(z)$

$\Rightarrow y=\frac54 x \mbox{ and }z=\frac32 x$

Substitute this into the first equation $x=\sqrt{\left(\frac54 x\right)^2-\frac{1}{16}}+\sqrt{\left(\frac32 x\right)^2-\frac{1}{16}}$ $4x=\sqrt{25x^2-1}+\sqrt{36x^2-1}$ $4x-\sqrt{25x^2-1}=\sqrt{36x^2-1}$ $41x^2-1-8x\sqrt{25x^2-1}=36x^2-1$ $5x^2=8x\sqrt{25x^2-1}$ $x=0$ is an extraneous solution. Dividing by $x$ and squaring both sides, $25x^2=1600x^2-64$ $x^2=\frac{64}{1575}$ $x=\frac{8}{15\sqrt7}$ $x+y+z=\frac{15}{4}x=\boxed{\dfrac{2}{\sqrt7}}$