Broken Triangle

Let a triangle with sides $a$ , $b$ , and $c$ have opposite angles be $\angle A$ , $\angle B$ , and $\angle C$ , respectively, and let it have an equalizer, a line that bisects the area and perimeter, that passes through sides $b$ and $c$ at points $D$ and $E$ , and let $b_1 = AD$ , $c_1 = AE$ , and $d = DE$ . Also let the perimeter of $\triangle ABC$ be $P = a + b + c$ .

Since $DE$ bisects the area, the area of $\triangle ADE$ is half the area of $\triangle ABC$ , so that $\frac{1}{2}b_1c_1 \sin A = \frac{1}{4}bc \sin A$ , which rearranges to $c_1 = \frac{bc}{2b_1}$ . Since $DE$ bisects the perimeter, the perimeter of $\triangle ADE$ is equal to the perimeter of quadrilateral $BCDE$ , so that $b_1 + c_1 + d = a + b - b_1 + d + c - c_1$ , which rearranges to $2b_1 + 2c_1 = a + b + c = P$ . Substituting $c_1 = \frac{bc}{2b_1}$ and solving for $b_1$ gives

$b_1 = \frac{1}{4}(P \pm \sqrt{P^2 - 8bc})$

Substituting this back into $c_1$ gives

$c_1 = \frac{1}{4}(P \mp \sqrt{P^2 - 8bc})$

Therefore, either $0$ , $1$ , or $2$ equalizers can exist through sides $b$ and $c$ depending on the value of the discriminant $P^2 - 8bc$ , and in order for the equalizer to exist, $0 \leq b_1 \leq b$ and $0 \leq c_1 \leq c$ .

Fortunately, both $b_1 > 0$ and $c_1 > 0$ no matter what:

$-8bc < 0$

$P^2 - 8bc < P^2$

$\sqrt{P^2 - 8bc} < P$

$0 < P - \sqrt{P^2 - 8bc}$

$0 < \frac{1}{4}(P - \sqrt{P^2 - 8bc})$

$0 < \frac{1}{4}(P - \sqrt{P^2 - 8bc}) < \frac{1}{4}(P + \sqrt{P^2 - 8bc})$

Now let's look at a right angle triangle. Let's say $\angle A$ is the right angle so that $a$ is the hypotenuse and $b$ and $c$ are the legs. Since $a > b$ and $a > c$ :

$a > b$

$a + b + c > 2b + c$

$P > 2b + c$

$-c > 2b - P$

$-8bc > 16b^2 - 8bP$

$P^2 - 8bc > 16b^2 - 8bP + P^2$

$+\sqrt{P^2 - 8bc} > 4b - P$

$\frac{1}{4}(P + \sqrt{P^2 - 8bc}) > b$

from the condition $b_1 \leq b$ above, $b_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc}) > b$ cannot be an equalizer.

By the same argument, if $a > c$ , then $c_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc})$ and cannot be an equalizer because then $c_1 > c$ .

Since either $b_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc})$ or $c_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc})$ , and neither can be equalizers, there can be no equalizers that pass through both legs of a right triangle:

Now let’s say $\angle A$ is an acute angle in a right triangle with legs $a$ and $b$ such that $a < b$ . Then assuming $P^2 - 8bc > 0$ :

$a < b$

$a + b + c < 2b + c$

$P < 2b + c$

$-c < 2b - P$

$-8bc < 16b^2 - 8bP$

$P^2 - 8bc < 16b^2 - 8bP + P^2$

$\pm \sqrt{P^2 - 8bc} < 4b - P$

$P \pm \sqrt{P^2 - 8bc} < 4b$

$\frac{1}{4}(P \pm \sqrt{P^2 - 8bc}) < b$

from the condition that $b_1 \leq b$ , it is possible for $b_1 = \frac{1}{4}(P \pm \sqrt{P^2 - 8bc}) < b$ to be equalizers.

By the arguments above there are two equalizers that pass through $b$ and $c$ , one for $b_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc})$ and $c_1 = \frac{1}{4}(P - \sqrt{P^2 - 8bc})$ , and one for $b_1 = \frac{1}{4}(P - \sqrt{P^2 - 8bc})$ and $c_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc})$ .

Now let’s say $\angle A$ is an acute angle in a right triangle with legs $a$ and $b$ such that $a > b$ . Then assuming $P^2 - 8bc > 0$ :

$a > b$

$a + b + c > 2b + c$

$P > 2b + c$

$-c > -P + 2b$

$-8bc > -8bP + 16b^2$

$P^2 - 8bc > P^2 - 8bP + 16b^2$

$+\sqrt{P^2 - 8bc} > P - 4b$

$4b > P - \sqrt{P^2 - 8bc}$

$b > \frac{1}{4}(P - \sqrt{P^2 - 8bc})$

$\frac{1}{4}(P - \sqrt{P^2 - 8bc}) < b$

from the condition $b_1 \leq b$ above, it is possible for $b_1 = \frac{1}{4}(P - \sqrt{P^2 - 8bc}) < b$ to be an equalizer.

However,

$a > b$

$a + b + c > 2b + c$

$P > 2b + c$

$-c > 2b - P$

$-8bc > 16b^2 - 8bP$

$P^2 - 8bc > 16b^2 - 8bP + P^2$

$+\sqrt{P^2 - 8bc} > 4b - P$

$\frac{1}{4}(P + \sqrt{P^2 - 8bc}) > b$

from the condition $b_1 \leq b$ above, $b_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc}) > b$ cannot be an equalizer.

By the arguments above there is only one equalizer that passes through $b$ and $c$ , for $b_1 = \frac{1}{4}(P - \sqrt{P^2 - 8bc})$ and $c_1 = \frac{1}{4}(P + \sqrt{P^2 - 8bc})$ .

Therefore in a right triangle, if $P^2 - 8bc > 0$ , there are $3$ equalizers: $0$ through both legs, $2$ through the larger leg and hypotenuse, and $1$ through the smaller leg and hypotenuse (and for an isosceles right triangle it’s still $3$ equalizers: $0$ through both legs, $1$ through the one leg and hypotenuse, and $1$ through the other leg and hypotenuse, and $1$ through the right angle and hypotenuse).

Therefore, for a right triangle with legs $a$ and $b$ to have exactly $2$ equalizers, $P^2 - 8bc \leq 0$ , which means:

$P^2 - 8bc \leq 0$

$P^2 \leq 8bc$

$(a + b + c)^2 \leq 8bc$

$a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \leq 8bc$

$c^2 + c^2 + 2ab + 2ac \leq 6bc$

$2c^2 + 2ab + 2ac \leq 6bc$

$c^2 + ab + ac \leq 3bc$

$\frac{c^2}{bc} + \frac{ab}{bc} + \frac{ac}{bc} \leq 3$

$\frac{c}{b} + \frac{a}{c} + \frac{a}{b} \leq 3$

$\sec A + \sin A + \tan A \leq 3$

This equation solves to approximately $\angle A \leq 43.267°$ , which means $a < b$ . If $\sec A + \sin A + \tan A < 3$ , then $P^2 - 8bc < 0$ , and $b_1 = \frac{1}{4}(P \pm \sqrt{P^2 - 8bc})$ and $c_1 = \frac{1}{4}(P \mp \sqrt{P^2 - 8bc})$ have no solutions, so there are no longer $2$ equalizers through the larger leg and hypotenuse but $0$ equalizers, reducing the number of equalizers down from $3$ to $1$ . But if $\sec A + \sin A + \tan A = 3$ , then $P^2 - 8bc = 0$ , and $b_1 = \frac{1}{4}(P \pm \sqrt{P^2 - 8bc})$ and $c_1 = \frac{1}{4}(P \mp \sqrt{P^2 - 8bc})$ have one solution each, so there are no longer $2$ equalizers through the larger leg and hypotenuse but $1$ equalizer, reducing the number of equalizers down from $3$ to $2$ , the desired amount.

Therefore, if $\tan A + \sin A + \sec A = \boxed{3}$ , there are exactly $2$ distinct ways of using a straight line to cut the regions so that they have the same area and same perimeter.

The lines that bisect the area and perimeter of a triangle simultaneously are called triangle equalizers .

Read the following papers to know more about equalizers:

• Lines simultaneously bisecting the perimeter and area of a triangle, Paul Yiu .

• Triangle Equalizers, Dimitrios Kodokostas .

• Equalizers in right angled triangles, Ioannis D. Sfikas .

Condition for one solution is (a+b+c)^2=8ab, where, a,b and c are sides of triangle with a^2+c^2=b^2. Choosing, a=1 and solve for c. Find the acute angle Z from value c. Calculate and find: Sec(Z)+Sin(Z)+Tan(Z)=3