1989 AHSME Problems/Problem 19

Revision as of 10:04, 20 May 2021 by Penguinempress (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?

$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }$


The three arcs make up the entire circle, so the circumference of the circle is $3+4+5=12$ and the radius is $\frac{12}{2\pi}=\frac{6}{\pi}$. Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as $3\theta$, $4\theta$, and $5\theta$ for some $\theta$. By Circle Angle Sum, we obtain $3\theta+4\theta+5\theta=360$. Solving yields $\theta=30$. Thus, the angles of the triangle are $90$, $120$, and $150$. Using $[ABC]=\frac{1}{2}ab\sin{C}$, we obtain $\frac{r^2}{2}(\sin{90}+\sin{120}+\sin{150})$. Substituting $\frac{6}{\pi}$ for $r$ and evaluating yields $\frac{9}{\pi^2}(\sqrt{3}+3)\implies{\boxed{E}}$.

-Solution by thecmd999

Solution 2 (no trig)

Since we know that the triangle is inscribed, we can use the properties of inscribed angles to solve the problem. At first, we can ignore actual lengths in this problem, and use the $3$, $4$, and $5$ as ratios instead of the actual numbers. There are $360$ degrees in a circle, so $3x + 4x + 5x = 360$, where we find $x$ to be $30$ degrees. Thus, the arc measures in degrees becomes $90$, $120$, and $150$. With this knowledge, the angle that cuts off the arc of length $3$ is $45$ degrees (an inscribed angle is $\frac{1}{2}$ the measure of the arc that it intercepts). Similarly, the angle cutting off length $4$ is $60$ degrees, and the last angle of the triangle is $75$ degrees.

Upon observing this figure, we can draw an altitude in the triangle, breaking it into two triangles, one being a 30-60-90 and the other being a 45-45-90. For now, we can set the leg of the 30-60-90 angle that is adjacent to the 60 degree angle as "x." This makes the other leg of the 30-60-90 degree angle to be x√3, and since that happens to be the altitude and also part of the 45-45-90 triangle, it makes both legs of the 45-45-90 triangle to be x√3. We sum together the areas of both triangles and get (x^2√3 + 3x^2)/2, or x^2(√3+3)/2.

Now it's time to actually find the value of x and plug it in. In order to find the value of x, we can use the length of the radius, which we identify as "r." Now, looking back at what we had earlier about the arcs, the arc of length 3 is 90 degrees, which is one-fourth of the circumference. The circumference itself is equal to 2rπ. So we can write this equation: 3 = 2rπ/4, making r = 6/π. Now we must find the relationship of r to x, but we don't have to look far to find it. In fact, if we draw the center of the circle and connect two radii from it to points of arc length 3, we see that it creates another 45-45-90 triangle, where r is the length of the leg, and as we see from earlier, the length of that hypotenuse is 2x. Since the hypotenuse is equal to the leg multiplied by √2 in such a triangle, we write this equation: 2x = 6√2/π, and x = 3√2/π. Now, we plug our new value of x into the equation x^2(√3+3)/2. Doing so, we get $\frac{9}{\pi^2}(\sqrt{3}+3)\implies{\boxed{E}}$.

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png