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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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#1 h Mar 23, 2025, 5:47 AM

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Equilateral $\triangle ADC$ is drawn externally on side $AC$ of $\triangle ABC$ . Point $P$ is taken on $BD$ . Find $\angle APC$ if $BD=PA+PB+PC$ .

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#2 h Mar 23, 2025, 6:22 AM

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Prove that $\triangle ABC$ is acute, right or obtuse according as $a^2+b^2+c^2-8R^2$ is positive, zero or negative.

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#3 h Monday at 2:59 PM

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Given $\triangle ABC\ :\ A(2,6),B(0,0),C(10,0)$ .
Point $D(6+3\sqrt{3},3+4\sqrt{3})$ .

Point $P(p,\frac{5\sqrt{3}-6}{3} \cdot p) \in BD$ .
Equality: $PD=AP+CP \Rightarrow p=1+\sqrt{3}$ , so $P(1+\sqrt{3},\frac{9-\sqrt{3}}{3})$ .
Then $\angle CPA=120^{\circ}$ .

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#4 h Monday at 3:39 PM

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JetFire008 wrote:

Equilateral $\triangle ADC$ is drawn externally on side $AC$ of $\triangle ABC$ . Point $P$ is taken on $BD$ . Find $\angle APC$ if $BD=PA+PB+PC$ .

Let $F$ be the Fermat point. Then $BD=FA+FB+FC$ , and $F$ is the unique point that minimizes this quantity. Thus $F=P$ . So $\angle APC=120^{\circ}$ .

JetFire008 wrote:

Prove that $\triangle ABC$ is acute, right or obtuse according as $a^2+b^2+c^2-8R^2$ is positive, zero or negative.

This is equivalent to comparing $\sin^2 A+\sin^2 B+\sin^2 C$ to $2$ . Recalling the identity
$\cos^2 A+\cos^2 B+\cos^2 C+2\cos A\cos B\cos C=1,$ we are done. $\blacksquare$

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