M∀TH Workout

May 2026

 
  • May 05, 2026

05/05/2026

$ABCD$ is a quadrilateral such that $\angle{ADB}=50^{\circ}$, $\angle{BDC}=30^{\circ}=\angle{ACB}$, $\angle{ACD}=40^{\circ}$, find $\angle{BAC}$

image-20260505014956285

Solve:

image-20260507202015383 \(\begin{multline} \shoveleft \text{Let }E \text{ be the circumcenter of }\triangle{BCD}, F=AC\cap BD\\ \shoveleft \implies EB=EC=ED, \angle{BEC}=2\angle{BCD}=60^{\circ}\\ \shoveleft \implies \triangle{BCE} \text{ is equilateral} \implies \\ \shoveleft \angle{ECF}=\angle{BCF}=30^{\circ}, \angle{DBE}=\angle{BDE}=20^{\circ} \implies \angle{ADE}=70^{\circ}\\ \shoveleft \implies \triangle{ABC}\cong\triangle{AEC},\triangle{BCF}\cong\triangle{ECF}\implies \angle{BAC}=\angle{EAC},BF=EF\\ \shoveleft \implies \angle{CFE}=180^{\circ}-30^{\circ}-60^{\circ}-20^{\circ}=70^{\circ}=\angle{ADE}\\ \shoveleft \implies ADEF \text{ is cyclic}\implies \angle{EAC}=\angle{BDE}=20^{\circ}\implies \angle{BAC}=\bbox[5px, border: 1px solid black]{20^{\circ}} \end{multline}\)

05/15/2026

$D$ is the midpoint of side $AC$ of $\triangle{ABC}$, and $\angle{ADB}=45^{\circ}$, $\angle{ACB}=30^{\circ}$, find $\angle{BAC}$

image-20260515001416402

Solve:

image-20260515001703517

\[\begin{multline} \shoveleft \text{Make }\triangle{ACE} \text{ to be an equilateral triangle}\\ \shoveleft \implies AC=AE=CE, \angle{ACE}=60^{\circ}\implies \angle{BCE}=30^{\circ}\\ \shoveleft AD=CD\implies DE\perp AC, \angle{AED}=\angle{CED}=30^{\circ}\\ \shoveleft \text{Let } F = AE\cap BC \implies AE \perp BC, AF=EF \implies \\ \shoveleft DF=AF=EF=AD=CD \implies \angle{DFC}=\angle{ACB}=30^{\circ}\\ \shoveleft \implies \angle{BDF}=\angle{CBD}=15^{\circ}\implies BF=DF=AF\\ \shoveleft \implies \angle{BAF}=45^{\circ}\implies \angle{BAC}=45^{\circ}+60^{\circ}=\bbox[5px, border: 1px solid black]{105^{\circ}} \end{multline}\]

05/16/2026

$B,C$ are on line $AD$ such that $\dfrac{AB}{BC}=\dfrac{AD}{CD}=k \ne 1$, $M$ is a point on the circle whose diameter is $BD$. Show that $\dfrac{AM}{MC}=k$

image-20260516221249830

Prove 1:

image-20260517024535851 \(\begin{multline} \shoveleft \text{Let }E \text{ on }AM \text{ such that }CE \parallel DM\\ \shoveleft \text{Let }F \text{ on }AB \text{ such that }EF \parallel MB\\ \shoveleft \text{Let }BC=a, DM=b, EM=c, BM=d, CM=x\\ \shoveleft \dfrac{AB}{a}=\dfrac{AB+a+CD}{CD}=k\implies AB=ka, CD=\dfrac{k+1}{k-1}a\\ \shoveleft \dfrac{AC}{CD}=k-1=\dfrac{AE}{c}\implies AE=(k-1)c\\ \shoveleft \dfrac{AE}{EM}=\dfrac{AF}{FB}=k-1, AF+FB=ka\implies AF=(k-1)a, FB=a\\ \shoveleft \dfrac{AE}{AM}=\dfrac{EF}{d}\implies EF=\dfrac{k-1}{k}d\\ \shoveleft \dfrac{AE}{AM}=\dfrac{EC}{b}\implies EC=\dfrac{k-1}{k}b\\ \shoveleft BD\text{ is the diameter }\implies BM\perp MD\implies \\ \shoveleft b^2+d^2=(1+\dfrac{k+1}{k-1})^2a^2=\dfrac{4k^2}{(k-1)^2}a^2 \implies d^2=\dfrac{4k^2}{(k-1)^2}a^2-b^2\\ \shoveleft \text{Apply Stewart's theorem on }\triangle{ACE} \text{ and }EF \implies\\ \shoveleft \dfrac{(k-1)^2}{k^2}b^2+2(k-1)c^2=\dfrac{(k^2-1)}{k^2}d^2+2(k+1)a^2\\ \shoveleft \implies \dfrac{k-1}{k}b^2+(k-1)c^2=\dfrac{(k+1)^2}{k-1}a^2\\ \shoveleft \text{Apply Stewart's theorem on }\triangle{ADM} \text{ and }CM \implies\\ \shoveleft (k+1)ab^2+\dfrac{k+1}{k-1}ak^2c^2=(k+1+\dfrac{k+1}{k-1})a(x^2+\dfrac{(k+1^2)}{k-1}a^2)\\ \shoveleft \implies x^2=\dfrac{k-1}{k}b^2+kc^2-\dfrac{(k+1)^2}{k-1}a^2=c^2\implies x=c \implies \dfrac{AM}{MC}=k\blacksquare\\ \end{multline}\) Prove 2:

image-20260517051352466 \(\begin{multline} \shoveleft \text{Make }PQ \perp BM \text{ such that }B \in PQ, AM\cap PQ=P, CM\cap PQ=Q \implies \\ \shoveleft \text{(1) }\triangle{BCQ}\sim\triangle{DCM}\implies\dfrac{BC}{CD}=\dfrac{BQ}{MD}\implies BQ=\dfrac{BC}{CD}MD\\ \shoveleft \text{(2) }\triangle{ABP}\sim\triangle{ADM}\implies\dfrac{AB}{AD}=\dfrac{PB}{MD}\implies PB=\dfrac{AB}{AD}MD\\ \shoveleft \dfrac{AB}{BC}=\dfrac{AD}{CD}=k \implies \dfrac{BC}{CD}=\dfrac{AB}{AD}\implies PB=BQ\implies \triangle{MPB}\cong\triangle{MQB}\\ \shoveleft \implies \angle{AMB}=\angle{CMB}\implies \dfrac{AM}{MC}=\dfrac{AB}{BC}=k\blacksquare \end{multline}\) Prove 3:

image-20260517061150072 \(\begin{multline} \shoveleft \text{Make }CP \parallel DM \text{ and } AM\cap CP=P, CQ\parallel BM \text{ and } AM \cap CQ=Q\\ \shoveleft \implies \dfrac{PM}{AM}=\dfrac{CD}{AD}, \dfrac{QM}{AM}=\dfrac{BC}{AB}\implies PM=\dfrac{CD}{AD}AM, QM=\dfrac{BC}{AB}AM\\ \shoveleft \dfrac{AB}{BC}=\dfrac{AD}{CD}=k \implies PM=QM\\ \shoveleft \angle{PCQ}=\angle{BMD}=90^{\circ}\implies M \text{ is the circumcenter of }\triangle{PCQ}\\ \shoveleft \implies QM=MC\implies \dfrac{AM}{MC}=\dfrac{AM}{QM}=\dfrac{AB}{BC}=k\blacksquare \end{multline}\) Note:

  1. The circle $(BDM)$ is the Apollonian Circle, which is the locus of points $M$ satisfying $\dfrac{AM}{MC}=k \ne 1$
  2. The circle $(BDM)$ inverts $A$ into $C$
  3. Any two of the following conditions implies the third condition (illustrated in EGMO by Evan Chen):
    • $(A, C; B, D)$ is a harmonic bundle (which means $\dfrac{AB}{CB} \cdot \dfrac{CD}{AD}=-1$, $-1$ means $\overline{AC}$ and $\overline{BD}$ intersect)
    • $M$ is a point not on the line and $\angle{BMD}=90^{\circ}$
    • $BM$ bisects $\angle{AMC}$

05/22/2026

$AC$ and $BD$ are two mutually perpendicular chords of a circle. $AB=6$, $CD=8$, find the radius of the circle.

image-20260522220906858

Solve 1:

image-20260522215028188 \(\begin{multline} \shoveleft \text{Easy to know }\triangle{ABE}\sim \triangle{DCE}, \text{Let }BE=3a,AE=3b \\ \shoveleft \implies CE=4a,DE=4b \implies BC=5a\\ \shoveleft \text{Let }AF\text{ be the diameter of the circle } \implies \angle{ACB}=\angle{AFB}\\ \shoveleft \implies sin\angle{ACB}=\dfrac{3}{5}=sin\angle{AFB}=\dfrac{6}{2R}\implies R=\bbox[5px, border: 1px solid black]{5} \end{multline}\) Solve 2: \(\begin{multline} \text {According to the }\href{https://math.stackexchange.com/questions/1481891/ab-and-cd-are-two-mutually-perpendicular-chords-of-a-circle-of-radius-r-i}{\text{anonymous theorem about two orthogonal chords}}\\ \shoveleft \implies AE^2+BE^2+CE^2+DE^2=4R^2\implies R=\sqrt{\dfrac{6^2+8^2}{4}}=\bbox[5px, border: 1px solid black]{5} \end{multline}\)

05/24/2026

$ABCD$ is a quadrilateral such that $\angle{ACB}=12^{\circ}$, $\angle{ACD}=57^{\circ}$, $\angle{ADB}=18^{\circ}$, $\angle{BDC}=30^{\circ}$, find $\angle{BAC}$

image-20260524015130396

Solve:

image-20260524015719833 \(\begin{multline} \shoveleft \text{Let }E \text{ be the circumcenter of }\triangle{BCD} \implies \\ \shoveleft BE=CE=BC, \angle{BEC}=2\angle{BDC}=60^{\circ}\\ \shoveleft \implies \triangle{BCE} \text{ is equilateral }\implies \angle{BCE}=60^{\circ}\\ \shoveleft \implies \angle{DCE}=9^{\circ}=\angle{EDC}\implies \angle{BDF}=21^{\circ},\angle{CEF}=18^{\circ}\\ \shoveleft \text{For this construction of equilateral triangle we proved }\href{https://mathwo.github.io/2024/05/01/wo/#05142024}{\text{HERE}}\\ \shoveleft \implies \angle{CBF}=6^{\circ}\implies \angle{BFA}=18^{\circ}=\angle{ADB}\\ \shoveleft \implies ABFD \text{ is cyclic}\implies \angle{BAC}=\angle{BDF}=\bbox[5px, border: 1px solid black]{21^{\circ}} \end{multline}\)

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