Category Archives: Geometry

Mathcounts 2017-2018 Handbook Problem 136

Posted on March 11, 2024 by kevin

In right triangle ABC, with AB = 44 cm and BC = 33 cm, point D lies on side BC so that BD:DC = 2:1. If vertex A is folded onto point D to create quadrilateral BCEF, as shown, 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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Geometry Challenge – 14

Posted on November 30, 2023 by kevin

In the figure as shown, $ABCD$ is a unit square, $E$ is on $BC$, $F$ is on $CD$, $\angle{EAF}=45^\circ$, $\angle{AEB}=70^\circ$. A B C D E F 70° 45° ? A B C D E F G 70° 45° ? (1) … 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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