M∀TH Workout

June 2025

 
  • Jun 12, 2025

06/12/2025

Consider a regular n-gon inscribed in the unit circle. Pick one vertex. Now connect it with line segments to all the other vertices. What is the value of the product of all the lengths of the segments?

Solve:

image-20250612155624929 \(\begin{multline} \shoveleft \text{Suppose }A_0, A_1, ... A_{n-1} \text{ are the vertices of this regular n-gon inscribed in the unit circle}\\ \shoveleft \text{Make }OD_k\perp A_0A_k \implies \angle{A_0OA_k}=\dfrac{k \cdot 2\pi}{n} \implies \angle{A_0OD_k=\dfrac{k\pi}{n}}, k=1, 2, ..., (n-1)\\ \shoveleft \implies A_0A_k=2 \cdot sin(\dfrac{k\pi}{n}) \implies ans=\displaystyle\prod_{i=1}^{n-1}{|A_0A_i|}=2^{n-1}\prod_{k=1}^{n-1}sin(\dfrac{k\pi}{n})\\ \shoveleft sin(\dfrac{k\pi}{n})=\dfrac{e^{i\tfrac{k\pi}{n}}-e^{-i\tfrac{k\pi}{n}}}{2i}\implies ans=2^{n-1}\prod_{k=1}^{n-1}{\dfrac{e^{i\tfrac{k\pi}{n}}-e^{-i\tfrac{k\pi}{n}}}{2i}}=\dfrac{2^{n-1}}{(2i)^{n-1}}\prod_{k=1}^{n-1}{(e^{i\tfrac{k\pi}{n}}-e^{-i\tfrac{k\pi}{n}})}\\ \shoveleft =\dfrac{1}{i^{n-1}}e^{-i(\tfrac{\pi+2\pi+...+(n-1)\pi}{n})}\prod_{k=1}^{n-1}{(e^{2i\tfrac{k\pi}{n}}-1)}=\dfrac{1}{i^{n-1}}e^{-i\tfrac{n(n-1)\pi}{2n}}\prod_{k=1}^{n-1}{(e^{2i\tfrac{k\pi}{n}}-1)}\\ \shoveleft i=e^{i\tfrac{\pi}{2}} \implies \dfrac{1}{i^{n-1}}e^{-i\tfrac{(n-1)\pi}{2}}=i^{-2(n-1)}=(-1)^{n-1}\implies ans=(-1)^{n-1}\prod_{k=1}^{n-1}{(e^{2i\tfrac{k\pi}{n}}-1)}=\prod_{k=1}^{n-1}{(1-e^{2i\tfrac{k\pi}{n}})}\\ \shoveleft 1, e^{2i\tfrac{k\pi}{n}}, k=1, 2, ..., n-1 \text{ are the roots of equation } z^n-1=0 \implies (z-1)\prod_{k=1}^{n-1}{(z-e^{2i\tfrac{k\pi}{n}})}=0=z^n-1\\ \shoveleft \implies \dfrac{z^n-1}{z-1}=\prod_{k=1}^{n-1}{(z-e^{2i\tfrac{k\pi}{n}})}=z^{n-1}+z^{n-2}+...+1 \implies \prod_{k=1}^{n-1}{(1-e^{2i\tfrac{k\pi}{n}})}=\sum_{k=0}^{n-1}{1^{k}}=n\implies ans=\bbox[5px, border: 1px solid black]{n} \end{multline}\)

06/15/2025

The function $f$ is defined by $f(x)=\dfrac{ax+b}{cx+d}$, where $a,b,c$ and $d$ are nonzero real numbers, has the properties $f(19)=19, f(97)=97$, and $f(f(x))=x$ for all values of $x$ except $-d/c$. Find the unique number that is not in the range of $f$.

Solve: \(\begin{multline} \shoveleft f(f(x))=\dfrac{a\dfrac{\dfrac{ax+b}{cx+d}}{}+b}{c\dfrac{ax+b}{cx+d}+d}=\dfrac{a^2x+ab+bcx+bd}{acx+bc+cdx+d^2}=\dfrac{(a^2+bc)x+b(a+d)}{c(a+d)x+(bc+d^2)}=x\\ \shoveleft \implies x\ne-\dfrac{bc+d^2}{c(a+d)}, -c(a+d)x^2+(a^2-d^2)x+b(a+d)=0\\ \shoveleft \implies c(a+d)=0, a^2=d^2, b(a+d)=0\\ \shoveleft f(19)=\dfrac{19a+b}{19c+d}=19 \implies 19a+b=19^2c+19d\\ \shoveleft f(97)=\dfrac{97a+b}{97c+d}=97 \implies 97a+b=97^2c+97d\\ \shoveleft \implies a-d=116c \ne 0 \implies a+d=0\\ \shoveleft \implies f(x)=\dfrac{ax+b}{cx-a}, a-d=2a=116c \implies \dfrac{a}{c}=58\implies a=58c\\ \shoveleft f(x)=\dfrac{ax+b}{cx-a}=\dfrac{58cx+b}{cx-58c}=58+\dfrac{b+58^2c}{c(x-58)}\\ \shoveleft \text{if }\dfrac{b+58^2c}{c(x-58)}=0 \implies f(x)=58 \implies f(19)=f(97)=58\\ \shoveleft \text{So }f(x)\ne \bbox[5px, border: 1px solid black]{58} \end{multline}\)

06/24/2025

$ABCD$ is a quadrilateral such that $AD=BC, \angle{ADC}=70^{\circ}, \angle{BCD}=60^{\circ}$, $E,F$ are the midpoint of $AB, CD$ respectively. Find $\angle{DFE}$.

image-20250625171457767

Solve:

image-20250625171809610 \(\begin{multline} \shoveleft \text{Let }M,N \text{ be the midpoint of }AC, BD \text{ respectively } \\ \shoveleft \implies EM\parallel BC \parallel NF, FM \parallel AD \parallel EN\\ \shoveleft FM=\dfrac{AD}{2}=EN=\dfrac{BC}{2}=EM=NF\\ \shoveleft \implies \angle{DFN}=60^{\circ}, \angle{CFM}=70^{\circ}, EMFN \text{ is a rhombus}\\ \shoveleft \implies\angle{EFN}=\angle{EFM}=\dfrac{180^{\circ}-60^{\circ}-70^{\circ}}{2}=25^{\circ}\\ \shoveleft \implies \angle{DFE}=\angle{DFN}+\angle{EFN}=60^{\circ}+25^{\circ}=\bbox[5px, border: 1px solid black]{85^{\circ}} \end{multline}\)

06/25/2025

$ABCD$ is a quadrilateral such that $\angle{ABD}=\angle{ADB}=18^{\circ}$, $\angle{CDB}=27^{\circ}$,$\angle{CBD}=45^{\circ}$ find $\angle{ACB}$.

image-20250625173559014

Solve 1:

image-20250625175751133 \(\begin{multline} \shoveleft \text{Make }E, F \text{ on }CD, BD \text{ respectively such that }AE\perp CD, AF\perp BD \\ \shoveleft \implies AFED \text{ is cyclic},\angle{ADE}=45^{\circ}\\ \shoveleft \implies \angle{DAE}=\angle{DFE}=45^{\circ}=\angle{CBD}\implies EF \parallel BC\\ \shoveleft \angle{ADB}=\angle{ABD}\implies DF=BF\implies DE=EC\\ \shoveleft \implies AD=AC=AB\implies \angle{ACB}=\angle{ABC}=\bbox[5px, border: 1px solid black]{63^{\circ}} \end{multline}\)

Solve 2:

image-20250627064512765 \(\begin{multline} \shoveleft \text{Let }E \text{ be a point on circumcircle of }\triangle{BCD}\\ \shoveleft \angle{BCD}=108^{\circ}\implies \angle{BED}=72^{\circ}\\ \shoveleft \angle{BAD}=144^{\circ}=2\angle{BED}, AB=AD\\ \shoveleft \implies A \text{ is circumcenter of }\triangle{BDE}\\ \shoveleft \implies A \text{ is circumcenter of }\triangle{BCD}\\ \shoveleft \implies AB=AC \implies \angle{ACB}=\angle{ABC}=\bbox[5px, border: 1px solid black]{63^{\circ}} \end{multline}\)

06/28/2025

2025 Youth Balkan Mathematics Olympic Problem 3: $\angle{BAC}=90^{\circ}$ in $\triangle{ABC}$ and $AD$ is the attitude of side $BC$. $E$ is the midpoint of $CD$. The circumcircle of $\triangle{ABD}$ meets $AE$ at $F$. Extended $BA$ and $DF$ meet at $X$. Show that $CX=DX$.

image-20250628233126170

Prove 1:

image-20250702163502832 \(\begin{multline} \shoveleft \text{Let } M \text{ be the midpoint of }AC, AD \cap BF=H, F \text{ is on }\odot{ABD} \\ \shoveleft \implies \angle{AFB}=\angle{ADB}=90^{\circ} \implies H \text{ is the orthocenter of }\triangle{ABE}\\ \shoveleft \implies EH \perp AB \implies EH \parallel AC, DE=CE \implies DH=AH\\ \shoveleft ABDF \text{ cyclic }\implies \triangle{ABH}\sim\triangle{FDH}, \triangle{BDH}\sim\triangle{AFH}, \triangle{XAF }\sim\triangle{XDB}\\ \shoveleft \implies \dfrac{XA}{DX}=\dfrac{AF}{BD}=\dfrac{FH}{DH}=\dfrac{FH}{AH}=\dfrac{DF}{AB}=\dfrac{FX}{BX}\implies \dfrac{DF}{FX}=\dfrac{AB}{BX}\\ \shoveleft \text{Apply Menelaus's Theorem on line }AFE \text{ and }\triangle{BDX} \implies \dfrac{DF}{FX}\dfrac{XA}{AB}\dfrac{BE}{DE}=1\\ \shoveleft \implies \dfrac{AB}{BX}\dfrac{XA}{AB}\dfrac{BE}{DE}=1\implies \dfrac{XA}{BX}=\dfrac{DE}{BE}=\dfrac{CE}{BE}\implies \dfrac{BE}{CE}\dfrac{CM}{MA}\dfrac{XA}{BX}=1\\ \shoveleft \implies AME \text{ are collinear by } \text{Menelaus's Theorem on }\triangle{ABC}\\ \shoveleft ME \parallel AD, AD \perp BC \implies XE \perp BC,DE=CE \implies XC=XD \blacksquare \end{multline}\) Prove 2:

image-20250703145524820 \(\begin{multline} \shoveleft \text{Let } M \text{ be the midpoint of }AC, AD \cap BF=H, F \text{ is on }\odot{ABD} \\ \shoveleft \implies \angle{AFB}=\angle{ADB}=90^{\circ} \implies H \text{ is the orthocenter of }\triangle{ABE}\\ \shoveleft \implies AFHG, DEFH, BDHG \text{ are all cyclic}\\ \shoveleft \implies EH \perp AB \implies EH \parallel AC, DE=CE \implies DH=AH=EM\\ \shoveleft \implies DEMH \text{ is rectangle }\implies DEMFH \text{ is cyclic}\implies XF\perp FM\\ \shoveleft \implies AFMX \text{ is cyclic }\implies \angle{FMX}=\angle{FAG}=\angle{EHF}=\angle{EDF}\\ \shoveleft =\angle{FDM}+\angle{EDM}=\angle{FEM}+\angle{EFM}\implies EMX \text{ is collinear}\\ \shoveleft ME \parallel AD, AD \perp BC \implies XE \perp BC,DE=CE \implies XC=XD \blacksquare \end{multline}\)

    © M∀TH Workout 2020-2021
    Search