Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.8 Junior Find the answer

The famous mathematician Qin Jiushao of the Southern Song Dynasty in China (approximately $1202-1261$ ) proposed the formula for calculating the area of a triangle using "three obliques to find the product." Subtract the middle oblique power from the product of the small oblique power and the large oblique power, take half of the result, multiply it by the upper part, multiply the small oblique power by the difference between the large oblique power and the upper part, take a quarter of the result, and it will be the actual area. One can find the product by taking the square root of the result obtained from the above description when written in a formula: $S=\sqrt{\frac{1}{4}[{{c^2}{a^2}-{{({\frac{{{c^2}+{a^2}-{b^2}}}{2}})}^2}}]}$ . In triangle $\triangle ABC$ , it is known that the sides opposite angles $A$ , $B$ , and $C$ are $a$ , $b$ , and $c$ , where $a$ and $c$ are the two roots of the equation $x^{2}-3x+2=0$ , and $B=\frac{π}{3}$ . What is the area of $\triangle ABC$ ?

Pick one

Solution

To solve the problem using the given formula and information, we follow these steps:

1. Find the roots of the given equation: The equation is $x^{2}-3x+2=0$ , which factors to $(x-2)(x-1)=0$ . Hence, the roots are $x=1$ and $x=2$ . These roots correspond to the lengths $a$ and $c$ of the triangle.

2. **Identify $a$ and $c$ :** Since $a$ and $c$ are the roots of the equation, we can assign $a=1$ and $c=2$ without loss of generality.

3. **Use the formula for the area of a triangle with sides $a$ , $b$ , and angle $B$ between $a$ and $c$ :** The formula for the area is given as $S=\frac{1}{2}ac\sin B$ . Given that $B=\frac{\pi}{3}$ , we substitute the values of $a$ , $c$ , and $B$ into the formula.

4. **Calculate the area $S$ :**
$S = \frac{1}{2} \times 2 \times 1 \times \sin\left(\frac{\pi}{3}\right) = \frac{1}{2} \times 2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

Therefore, the area of triangle $\triangle ABC$ is $\boxed{\frac{\sqrt{3}}{2}}$ , which corresponds to option C.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.