Tricks to consider: since there are 3 values (sides or angles), it is often useful to write the sequence symmetrically, i.e. $a-d, a, a+d$ rather than $a,a+d,a+2d$. This helps with cancellation, and for angles, you avoid having double angles.
Notice that the sum of the angles is $180^\circ$ in a triangle. This implies that $(a-d)+a+(a+d)=3a=180^\circ$, so that $a=60^\circ$. Hence, the middle term for angles is always $60^\circ$, and you can write the sequence as $60^\circ-d, 60^\circ, 60^\circ+d$.
The largest side is opposite the largest angle and the smallest side is opposite the smallest angle.
2013 AMC 12A Problems/Problem 12
The angles in a particular triangle are in arithmetic progression, and the side lengths are $4,5,x$. The sum of the possible values of $x$ equals $a+\sqrt{b}+\sqrt{c}$ where $a, b$, and $c$ are positive integers. What is $a+b+c$?
Solution:
Let the angles be $60^\circ-d, 60^\circ, 60^\circ+d$. We notice that we have one angle and have to relate all sides, so this is a hint to use the cosine law. Since the cosine law is symmetric in the two legs, we apply casework on $x$.
Cases on $x$:
- $x$ is the hypotenuse.
- $x$ is a leg.
Pre-computations: $\cos 60^\circ = \frac{1}{2}$.
Case 1: $x$ is the hypotenuse. $x^2 = 4^2 + 5^2 - 2\cdot4\cdot5\cdot\cos 60^\circ$ $x = \sqrt{21}$.
Case 2: $x$ is a leg. $5^2 = 4^2 + x^2 - 2\cdot4\cdot x\cdot\cos 60^\circ$ $0 = x^2 - 4x - 9$ $13 = x^2-4x+4=(x-2)^2$ $x=2+\sqrt{13}$.
Hence the total sum is $2+\sqrt{13}+\sqrt{21}$, giving $2+13+21=\boxed{36}$.
2023 AMC 12B Problems/Problem 17
Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6$. If the triangle has an angle of $120^\circ$, what is the area of $ABC$?
Again, 3 sides and 1 angle, hinting toward cosine law. Let the sides be $6,6+d,6+2d$. By the cosine law with $\cos 120^\circ=-\frac{1}{2}$:
$$(6+2d)^2=6^2+(6+d)^2-2\cdot6\cdot(6+d)\cdot\cos 120^\circ$$ $$36+24d+4d^2=36+(36+12d+d^2)+(36+6d)$$ $$3d^2+6d-72=0$$ $$d^2+2d-24=0$$ $$(d+6)(d-4)=0$$Hence $d=4$. Since $120^\circ \ge 90^\circ$, it must be opposite to the hypotenuse $6+2d$, so the angle is included between the legs. By the sine area formula, we have $\text{Area} = \frac{1}{2}6\cdot(6+4)\sin 120^\circ=30 \cdot \frac{\sqrt{3}}{2}=\boxed{15\sqrt{3}}$.
2022 Euclid Problem 8(b)
Consider the following statement:
There is a triangle that is not equilateral whose side lengths form a geometric sequence, and the measures of whose angles form an arithmetic sequence.
Show that this statement is true by finding such a triangle or prove that it is false by demonstrating that there cannot be such a triangle.
Solution:
Some potential paths involve showing that if the triangle is not equilateral, then the triangle inequality is violated, or some sides are negative.
Let the angles be $60^\circ-d<60^\circ<60^\circ+d$. Let the side lengths be $\frac{1}{a}<1<a$. (We can scale to a similar triangle.)
We have information about all sides and angles, and we want to relate them all. This is a hint to use the sine law.
$$\frac{1}{a\sin{60^\circ-d}}=\frac{1}{\sin{60^\circ}}=\frac{a}{\sin{60^\circ+d}}$$We can isolate $a$ in the above equations to eliminate it:
$$a = \frac{\sin{60^\circ}}{\sin{60^\circ-d}}$$and
$$a=\frac{\sin{60^\circ+d}}{\sin{60^\circ}}$$Equating the two:
$$\sin(60^\circ-d)\sin(60^\circ+d)=\sin^2{60^\circ}=\frac{3}{4}$$Using the addition formulas:
$$\sin{60^\circ-d}=\sin{60^\circ}\cos{d}-\cos{60^\circ}\sin{d}$$ $$=\frac{\sqrt{3}}{2}\cos d - \frac{1}{2}\sin{d}$$ $$\sin{60^\circ+d}=\frac{\sqrt{3}}{2}\cos d + \frac{1}{2}\sin{d}$$Hence:
$$\frac{3}{4}\cos^2 d - \frac{1}{4}\sin^2{d}=\frac{3}{4}$$ $$3\cos^2 d - \sin^2{d}=3$$Substitute $\sin^2{d}=1-\cos^2{d}$:
$$3\cos^2 d - 1+\cos^2{d}=3$$ $$\cos^2{d} = 1$$ $$\cos{d} = \pm 1$$The only solutions are multiples of $180^\circ$, so the only valid case in triangles would be $d=0^\circ$, but that means it is equilateral. Proof complete.