2022-04-04
In $\triangle{ABC}$, $\angle{ABC}=48^{\circ}, \angle{ACB}=30^{\circ}$. Extend $BC$ t0 $D$ so that $AB=CD$. Find $\angle{ADC}$.
Solve: \(\begin{multline} \shoveleft \text{Make } F \text{ on } BC \text{ so that } \angle{AFB}=\angle{ABF}=48^{\circ}. \\ \shoveleft \text{ Make equilateral triangle } \triangle{AGF}. \text{ Connect } BG, CG, DG. \text { Easy to see} \\ \shoveleft AB=AF=AG=FG, \angle{AGF}=60^{\circ}=2 \cdot \angle{ACF}, AG=FG\\ \shoveleft \implies G \text{ is the circumcenter of } \triangle{ACF} \\ \shoveleft \implies CG=FG=AG=AF=AB \\ \shoveleft \angle{GFC}=180^{\circ}-48^{\circ}-60^{\circ}=72^{\circ} \\ \shoveleft \implies \angle{FGC}=36^{\circ}, \angle{FCG}=72^{\circ} \\ \shoveleft \implies \angle{CDG}=\angle{CGD}=36^{\circ} \\ \shoveleft \implies \angle{FGD}=72^{\circ}=\angle{GFD} \\ \shoveleft \implies \triangle{ADG} \cong \triangle{ADF} \\ \shoveleft \implies \angle{ADC}=\dfrac{\angle{FDG}}{2}=\bbox[1px, border: 1px solid black]{18^{\circ}} \nonumber \end{multline}\)
2022-04-10
Solve the equation group in real numbers:
\(\begin{multline} \begin{cases} x^3+2x-y=1 \\ y^3+y+x=11 \end{cases} \nonumber \end{multline}\)
Solve: \(\begin{multline} \shoveleft x^2+2x-y=1 \implies x^3+2x-3=y-2 \\ \shoveleft \implies (x-1)(x^2+x+3)=y-2 \\ \shoveleft x \in \mathbb{R} \implies x^2+x+3 > 0 \\ \shoveleft \implies (x-1)(y-2) \ge 0 \\ \shoveleft y^3+y+x=11 \implies y^3+y-10=-(x-1) \\ \shoveleft \implies (y-2)(y^2+2y+5)=-(x-1) \\ \shoveleft y \in \mathbb{R} \implies y^2+2y+5 > 0 \\ \shoveleft \implies (x-1)(y-2) \le 0 \\ \shoveleft \text{So we know }\bbox[1px, border: 1px solid black]{x=1, y=2} \nonumber \end{multline}\)
2022-04-11
$D$ is a point in $\triangle{ABC}$ so that $\angle{ABD}=10^{\circ}, \angle{DBC}=20^{\circ}, \angle{BCD}=40^{\circ}, \angle{DAC}=50^{\circ}$, find $\angle{BAD}$.
Solve 1:
This problem is easily done with trigonometric Ceva theorem as: \(\begin{multline} \shoveleft \dfrac{sinx}{sin50^{\circ}}\dfrac{sin(60^{\circ}-x)}{sin40^{\circ}}\dfrac{sin20^{\circ}}{sin10^{\circ}}=1 \\ \shoveleft \implies sinx \cdot sin(60^{\circ}-x)=\dfrac{sin50^{\circ}\cdot sin40^{\circ}\cdot sin10^{\circ}}{sin20^{\circ}} \\ \shoveleft =\dfrac{cos40^{\circ} \cdot sin40^{\circ} \cdot sin10^{\circ}}{2sin10^{\circ}\cdot cos10^{\circ}}=\dfrac{sin80^{\circ}}{4cos10^{\circ}}=\dfrac{1}{4} \\ \shoveleft \implies -\dfrac{cos60^{\circ}-cos(2x-60^{\circ})}{2}=\dfrac{1}{4} \\ \shoveleft \implies cos(2x-60^{\circ})=1 \\ \shoveleft \implies 2x-60^{\circ}=0 \\ \shoveleft \implies x=\bbox[1px, border: 1px solid black]{30^{\circ}} \end{multline}\)
Solve 2:
Here is second solution: \(\begin{multline} \shoveleft \text{Let } E \text{ be the circumcenter of } \triangle{ACD}, so \\ \shoveleft AE=CE=DE, \angle{DEC}=2\angle{DAC}=100^{\circ} \\ \shoveleft \implies \angle{EDC}=\angle{ECD}=40^{\circ}=\angle{BCD} \\ \shoveleft \implies DE \parallel BC \\ \shoveleft \text{Extend } ED \text{ and intersects with } AB \text{ at }F, \text{so}\\ \shoveleft \angle{AFE}=\angle{ABC}=30^{\circ}. \text{Make } G \text{ on } BC \text{ so that}\\ \shoveleft \angle{EFG}=100^{\circ}=\angle{FEC} \\ \shoveleft \implies CEFG \text{ is an isosceles trapezoid } \\ \shoveleft \implies FG=CE\\ \shoveleft \text{Let } H \text{ be the circumcenter of } \triangle{BFG}, \text{ so}\\ \shoveleft BH=GH=FH, \angle{FHG}=2\angle{FBG}=60^{\circ} \\ \shoveleft \implies \triangle{FGH} \text{ is an equilateral triangle} \\ \shoveleft \implies FH=GH=BH=FG=CE=DE=AE, \angle{HFG}=60^{\circ} \\ \shoveleft \implies \angle{HBF}=\angle{HFB}=\angle{HFG}-\angle{BFG}=\angle{HFG}-(180^{\circ}-\angle{FBG}-\angle{FGB}) \\ \shoveleft =\angle{HFG}-(180^{\circ}-\angle{FBG}-\angle{GFE}) =60^{\circ}-(180^{\circ}-100^{\circ}-30^{\circ})=10^{\circ} \\ \shoveleft \implies FH \parallel BD, \angle{HBD}=20^{\circ}=\angle{DBC}=\angle{FDB} \\ \shoveleft \implies BDFH \text{ is an isosceles trapezoid} \\ \shoveleft \implies BH=DF \implies DF=DE=AE \\ \shoveleft \text{Check }\triangle{AEF}, \angle{AFE}=30^{\circ}, AE=DE=DF \\ \shoveleft \implies EF=2AE \\ \shoveleft \text{Let } O \text{ be the circumcenter of } \triangle{AEF}, \text{ so}\\ \shoveleft \angle{AOE}=2 \angle{AFE}=60^{\circ}, AO=FO=EO \\ \shoveleft \implies \triangle{AOE} \text{ is equilateral triangle} \\ \shoveleft \implies AO=OE=AE=DE=DF=FO=EO \\ \shoveleft \implies AD=AO=FD \\ \shoveleft \implies \angle{BAD}=\angle{AFE}=\bbox[1px, border: 1px solid black]{30^{\circ}} \end{multline}\)
Solve 3:
\(\begin{multline}
\shoveleft \text{This approach is given by Stan Fulger at } \href{math.stackexchange.com/questions/4423750/need-pure-geometric-solution-for-proof-on-10-20-40-50-angle-problem/5080009#5080009}{\text{HERE}}\\
\shoveleft \text{'Double Angle Lemma: '}\\
\shoveleft \text{If } D \text{ is a point on altitude } BG \text{ of }\triangle{ABC} \text{ such that }\\
\shoveleft \angle{BAD}=30^{\circ}, \angle{CBD} + 30^{\circ}=\angle{ABD},\text{ then }\angle{BCD}=2\angle{CBD}\\
\end{multline}\)
\(\begin{multline}
\shoveleft \text{To prove this lemma, let } \angle{CBD}=\alpha\\
\shoveleft OH \text{ is the permendicular pisector of }AC, AC \cap BC=O\\
\shoveleft \implies OA=OC, \angle{AOH}=\angle{COH}=\alpha, \angle{BAO}+\angle{AOB}=\angle{ABC}\\
\shoveleft \implies \angle{OAB}=30^{\circ}=\angle{BAD}\\
\shoveleft \text{Let }E \text{ be the reflection of }O \text{ along }AB\\
\shoveleft \implies OA=EA=OC, \angle{ABB}=\angle{AOB}=2\alpha\\
\shoveleft \implies O \text{ is the circumcenter of }\triangle{ACE}\\
\shoveleft \implies \angle{AEC}=\dfrac{\angle{AOC}}{2}=\alpha=\angle{CBD} \implies BDCE \text{ is cyclic}\\
\shoveleft \implies \angle{BCD}=\angle{BED}=2\alpha =2\angle{CBD}\blacksquare
\end{multline}\)
2022-04-19
Point $D$ is in $\triangle{ABC}$ so that $\angle{ABD}=5^{\circ}, \angle{DBC}=20^{\circ}, \angle{DCB}=65^{\circ}, \angle{DAC}=40^{\circ}$, find $\angle{ACD}$.
Solve 1:
Trigonometric Ceva theorem approach is quite straight forward: \(\begin{multline} \shoveleft \dfrac{sin(50^{\circ}-x)}{sin40^{\circ}}\dfrac{sinx}{sin65^{\circ}}\dfrac{sin20^{\circ}}{sin5^{\circ}}=1 \\ \shoveleft \implies sin(50^{\circ}-x)sinx=\dfrac{sin5^{\circ} \cdot sin40^{\circ}\cdot sin65^{\circ}}{sin20^{\circ}}=2sin5^{\circ}\cdot cos20^{\circ} \cdot sin65^{\circ}\\ \shoveleft \implies cos(50^{\circ}-2x)-cos50^{\circ}=4sin5^{\circ} \cdot cos20^{\circ} \cdot cos 25^{\circ} \\ \shoveleft \implies cos(50^{\circ}-2x)=4sin5^{\circ} \cdot cos20^{\circ} \cdot cos 25^{\circ}+sin40^{\circ}\\ \shoveleft \qquad =4sin5^{\circ} \cdot cos20^{\circ} \cdot cos 25^{\circ}+2sin20^{\circ}\cdot cos20^{\circ}\\ \shoveleft \qquad =2cos20^{\circ}(2sin5^{\circ}sin65^{\circ}+cos70^{\circ})\\ \shoveleft \qquad =2cos20^{\circ}(cos60^{\circ}-cos70^{\circ}+cos70^{\circ})\\ \shoveleft \qquad =cos20^{\circ}\\ \shoveleft \implies 50^{\circ}-2x=\pm 20^{\circ} \\ \shoveleft \implies x=\bbox[1px, border: 1px solid black]{15^{\circ}} \text{ or } x=\bbox[1px, border: 1px solid black]{35^{\circ}} \nonumber \end{multline}\) When $x=35^{\circ}, \angle{ACB} > 90^{\circ}$:
Solve 2: