August 2025

 

08/13/2025

$\triangle{ABC}$ is inscribed in a circle and the tangent lines at $A,B$ meet at $M$, the tangent lines at $B,C$ meet at $N$. $BP\perp AC$ at $P$, show that $\angle{MPB}=\angle{NPB}$.

image-20250813134044655

Prove 1:

image-20250813135107226 \(\begin{multline} \shoveleft \text{Let }E, F \text{ on extended }AC \text{ such that }EM\perp MN, FN \perp MN\\ \shoveleft \implies EMBP, FNBP \text{ is both cyclic }\\ \shoveleft \implies \angle{MPB}=\angle{MEB}, \angle{NBP}=\angle{NFB}\\ \shoveleft \text{Let }MB=MA=a, NB=NC=b\\ \shoveleft \angle{MAB}=\angle{MBA}=\angle{BCA}=\alpha, \angle{NBC}=\angle{NCB}=\angle{BAC}=\beta\\ \shoveleft \implies sin\angle{AEM}=sin\angle{NBP}=sin\angle{NFC}\\ \shoveleft \implies \angle{MAE}=180^{\circ}-\alpha-\beta=\angle{NCF} \implies sin\angle{MAE}=sin\angle{NCF}\\ \shoveleft \implies \dfrac{ME}{MB}=\dfrac{ME}{MA}=\dfrac{sin\angle{MAE}}{sin\angle{AEM}}=\dfrac{sin\angle{NCF}}{sin\angle{NFC}}=\dfrac{NF}{NC}=\dfrac{NF}{NB}\\ \shoveleft \implies \triangle{EMB}\sim\triangle{FNB}\implies \angle{MEB}=\angle{NFB}\implies \angle{MPB}=\angle{NPB} \blacksquare \end{multline}\) Prove 2:

image-20250814134831210 \(\begin{multline} \shoveleft \text{Let }E, F \text{ on extended }AC \text{ such that }EM\perp MN, FN \perp MN\\ \shoveleft \implies EMBP, FNBP \text{ is both cyclic }\\ \shoveleft \implies \angle{MPB}=\angle{MEB}, \angle{NBP}=\angle{NFB}\\ \shoveleft \text{Let }D \text{ on }EF \text{ such that }ND=NF\implies \angle{CDN}=180^{\circ}-\angle{NFD}\\ \shoveleft \implies \angle{MEA}=\angle{NBP}=180^{\circ}-\angle{NFD}=\angle{CDN}\\ \shoveleft \angle{MAB}=\angle{MBA}=\angle{BCA}, \angle{NBC}=\angle{NCB}=\angle{BAC}\\ \shoveleft \implies \angle{MAE}=\angle{NCF} \implies \triangle{AEM}\sim\triangle{CDN}\\ \shoveleft \implies \dfrac{ME}{MA}=\dfrac{ND}{NC}\implies \dfrac{ME}{MB}=\dfrac{NF}{NB} \implies \triangle{EMB}\sim\triangle{FNB}\\ \shoveleft \implies \angle{MEB}=\angle{NFB} \implies \angle{MPB}=\angle{NPB} \blacksquare \end{multline}\)

Prove 3:

image-20250815014906432 \(\begin{multline} \shoveleft \text{Let }O \text{ be the center of the circumcircle of }\triangle{ABC}\\ \shoveleft \implies OB\perp MN, MO\perp AB, NO \perp BC\\ \shoveleft \implies \angle{ABM}=\angle{ACB}=\angle{MOB}, \angle{CAB}=\angle{CBN}=\angle{NOB}\\ \shoveleft \implies \triangle{BMO}\sim\triangle{PBC}, \triangle{BNO}\sim\triangle{PBA}\\ \shoveleft \implies \dfrac{MB}{BO}=\dfrac{BP}{PC},\dfrac{NB}{BO}=\dfrac{BP}{PA}\implies \dfrac{MB}{NB}=\dfrac{PA}{PC}=\dfrac{MA}{NC}\\ \shoveleft \angle{PAM}=\angle{PCN}\implies \triangle{MAP}\sim\triangle{NCP}\\ \shoveleft \implies \angle{MPA}=\angle{NPC} \implies \angle{MPB}=\angle{NPB}\blacksquare \end{multline}\)


08/18/2025

$ABCD$ and $A’B’C’D’$ are both squares and $D’A’ \cap AB = E$, $A’B’\cap BC=F$,$B’C’ \cap CD = G$, $C’D’ \cap DA = H$. Show that $EG\perp FH$

image-20250818142345238

Prove:

image-20250818145914573 \(\begin{multline} \shoveleft \text{Let }A'D'\cap CD = E', D'C' \cap BC=H', EG\cap FH=O\\ \shoveleft E'M \perp AB \text{ at } M, H'N' \perp DA \text{ at }N\\ \shoveleft GP \perp EE' \text{ at }P, FQ \perp HH' \text{ at }Q\\ \shoveleft AD\perp CD, D'H\perp D'E' \implies D'E'DH \text{ is cyclic }\\ \shoveleft \implies \angle{EE'G}=\angle{NHH'}=\angle{MEE'}=\angle{FH'H}\\ \shoveleft NH'=ME'=AB \implies \triangle{HNH'}\cong\triangle{EME'}\\ \shoveleft \implies EE'=HH'\\ \shoveleft PG=C'D'=B'C'=FQ \implies \triangle{PE'G}\cong\triangle{H'QF}\\ \shoveleft \implies E'G=H'F\implies \triangle{FHH'}\cong\triangle{E'EG}\\ \shoveleft \implies \angle{H'HF}=\angle{E'EG}\implies D'HEO \text{ is cyclic}\\ \shoveleft \implies \angle{EOH}=\angle{ED'H}=90^{\circ}\implies EG \perp FH \blacksquare \end{multline}\)


08/22/2025

Find the area of the square:

image-20250822121841779

Solve:

image-20250822122159476 \(\begin{multline} \shoveleft \text{Let }BD \cap CE=M, [BEH]=x,[BHM]=y\\ \shoveleft \implies [BCE]=[BDE]\implies [DGHM]=6-y\\ \shoveleft [BEC]+[ADE]=[CDE]\implies[CDGH]=18+x\\ \shoveleft \implies [CDM]=12+x+y\\ \shoveleft \text{Apply Cross Ladder Theorem on }\triangle{ABD} \implies\\ \shoveleft \dfrac{1}{[BDE]}+\dfrac{1}{[BDF]}=\dfrac{1}{[BDG]}+\dfrac{1}{[BDA]}\\ \shoveleft \implies \dfrac{1}{8}+\dfrac{1}{x+9}=\dfrac{1}{6}+\dfrac{1}{x+21} \implies x=3\\ \shoveleft \implies [ABCD]=2(21+x)=\bbox[5px, border: 1px solid black]{48} \end{multline}\)


08/27/2025

$ABCD$ is a square and $E$ is inside it such that $CEGF$ form another square and $AE \parallel DG$. Show that $GD\perp DE$.

image-20250827114616839

Prove:

image-20250827114824895 \(\begin{multline} \shoveleft \text{Extend } DG \text{ to }G' \text{ such that }AEGG' \text{ form a parallelogram}\\ \shoveleft \text{Make }AB'F'G' \text{ to be a square} \implies AB'\parallel CE\implies \angle{ACE}=\angle{B'AC}\\ \shoveleft \angle{ACD}=\angle{ECG}=45^{\circ}\implies \angle{DCG}=\angle{ACE}=\angle{B'AC}\\ \shoveleft \implies \angle{DAG'}=90^{\circ}-45^{\circ}-\angle{B'AC}=90^{\circ}-45^{\circ}=\angle{DCG}=\angle{DCE}\\ \shoveleft AG'=CE, AD=CD\implies \triangle{ADG'}\cong\triangle{CDE}\implies \angle{ADG'}=\angle{CDE}\\ \shoveleft \implies \angle{CDE}+\angle{ADE}=90^{\circ}=\angle{ADG'}+\angle{ADE}=\angle{EDG'}\implies GD\perp DE \blacksquare\\ \end{multline}\)


test some code change