Category Archives: Trigonometry

A MATHCOUNTS Problem with Multiple Answers 🙂

Posted on August 19, 2024 by kevin

The following is MATHCOUNTS 2012 National Competition Sprint Round Problem 22: In circle $O$, shown, $OP=2$ units, $PL=8$ units, $PK=9$ units and $NK=18$ units. Points $K$, $P$ and $M$ are collinear, as are points $L$, $P$, $O$ and $N$. What … Continue reading →

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Geometry Challenge – 15

Posted on March 3, 2024 by kevin

In $\triangle{ABC}$, $D$ is a point on $BC$. $\angle{ABC}=100^\circ$, $\angle{BCA}=20^\circ$, $\angle{BAD}=50^\circ$. Prove $AB=CD$. Click here for the proof. Proof 1: Clearly, $\angle{ADC}=150^\circ$, and $\angle{CAD}=10^\circ$. According to the law of sines, we have: $ \dfrac{sin\angle{BCA}}{AB}=\dfrac{sin\angle{ABC}}{AC}$ and $ \dfrac{sin\angle{CAD}}{CD}=\dfrac{sin\angle{ADC}}{AC}$ Therefore $AB=\dfrac{sin\angle{BCA}}{sin\angle{ABC}}AC=\dfrac{sin(20^\circ)}{sin(100^\circ)}AC=\dfrac{2sin(10^\circ)cos(10^\circ)}{sin(80^\circ)}AC=\dfrac{2sin(10^\circ)cos(10^\circ)}{cos(10^\circ)}AC$ $$=2sin(10^\circ)AC$$ … Continue reading →

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Trigonometry Challenge – 1 ⭐⭐

Posted on May 13, 2023 by kevin

Prove that $\ \ \ \ \ \ \ \dfrac{sin 20^\circ}{cos 20^\circ-2\cdot sin 10^\circ}=tan 30^\circ\ \ \ \ \ \ \ \ \ $ Solution Proof: Because $$ sin 10^\circ-sin 10^\circ=0,\ \ \ sin^2 20^\circ+cos^2 20^\circ=1,\ \ \ sin 30^\circ … Continue reading →

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Geometry Challenge – 13

Posted on April 18, 2023 by kevin

$ABCD$ is a square, P is an inner point such that $PA:PB:PC=1:2:3$. Find $\angle{APB}$ in degrees. A B C D P Click here for the solution. Solution 1: As shown in the diagram at the right, link $AC$. Without loss … Continue reading →

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AIME 2022 II – Problem 15

Posted on February 3, 2023 by kevin

Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = … Continue reading →

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