04/05/2026
$AB $ is the diameter of a circle and $AB =8$. $P$ is a point on the circle. Find the maximum value of $PA+PB$.
Solve:
\(\begin{multline}
\shoveleft \text{Extend }AP \text{ to } Q \text{ such that }PB=PQ\\
\shoveleft \implies \dfrac{AB}{sin(45^{\circ})}=\dfrac{AQ}{sin(45^{\circ}+\angle{ABP})}\\
\shoveleft \implies AQ=\dfrac{AB}{sin(45^{\circ})}sin(45^{\circ}+\angle{ABP})\\
\shoveleft \implies AQ \le \dfrac{AB}{sin(45^{\circ})}=\bbox[5px, border: 1px solid black]{8\sqrt{2}}
\end{multline}\)
04/11/2026
$\triangle{ABC}$ is an isosceles with $AB=AC$, $D$ is a point inside it such that $AD=BC$, $\angle{ABD}=\alpha$, $\angle{CAD}=\angle{BCD}=x$, $\angle{ACD}=2\alpha$, $\angle{CBD}=x+\alpha$, find $x, \alpha$
Solve:
\(\begin{multline}
\shoveleft \text{Easy to see }\angle{BAD}=180^{\circ}-2\angle{ABC}-\angle{CAD}=180^{\circ}-4\alpha-3x\\
\shoveleft \text{Extend }CD \text{ to } E,F \text{ such that }AE=AD=BC, AF=AC=AB\\
\shoveleft \implies \angle{AFD}=\angle{ACD}=2\alpha, \angle{EAF}=\angle{CAD}=x, \angle{BAF}=2x \\
\shoveleft \implies \angle{BAE}=x=\angle{FAE}=\angle{BCD}\implies\\
\shoveleft \text{(1) } AEBC \text{ is cyclic} \implies \\
\shoveleft \angle{ABE}=\angle{ACE}=2\alpha, \angle{BAC}=\angle{ACE} \implies \angle{EAC}=\angle{BCA} \\
\shoveleft \implies \triangle{BAC}\cong\triangle{ECA}\implies AB=AC=EC=DF\\
\shoveleft \text{(2) }\triangle{FAE}\cong\triangle{BAE} \implies FE=EB \implies \angle{EFB}=\angle{EBF}=\alpha\\
\shoveleft \implies F\text{ is circumcenter of }\triangle{BAD}\implies BF=AF=DF\\
\shoveleft \implies \triangle{ABF}\text{ is equilateral}\implies 2x=3\alpha=60^{\circ}\implies x=\bbox[5px, border: 1px solid black]{30^{\circ}},\alpha=\bbox[5px, border: 1px solid black]{20^{\circ}}
\end{multline}\)