Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Dao Thanh Oai geometric results"

(→‎Quadrilateral with side bisectors)
(→‎Quadrilateral with side bisectors)
Line 5: Line 5:
 
==Quadrilateral with side bisectors==
 
==Quadrilateral with side bisectors==
 
[[File:Dao ginma1.png|300px|right]]
 
[[File:Dao ginma1.png|300px|right]]
 +
[[File:Dao ginma2.png|300px|right]]
 
Let a convex quadrilateral <math>ABCD</math> be given. Let <math>\ell', M'</math> and <math>\ell, M</math> be the bisector and the midpoint  of <math>AB</math> and <math>CD,</math> respectively. Let <math>\ell'</math> intersect <math>\ell</math> at the point <math>P</math> inside <math>ABCD.</math> Denote <math>\angle APM' =  \alpha, \angle MPD = \beta.</math>
 
Let a convex quadrilateral <math>ABCD</math> be given. Let <math>\ell', M'</math> and <math>\ell, M</math> be the bisector and the midpoint  of <math>AB</math> and <math>CD,</math> respectively. Let <math>\ell'</math> intersect <math>\ell</math> at the point <math>P</math> inside <math>ABCD.</math> Denote <math>\angle APM' =  \alpha, \angle MPD = \beta.</math>
 
   
 
   

Revision as of 05:50, 29 August 2025

Dao Thanh Oai was born in Vietnam in 1986. He is an engineer with many innovative solutions for Vietnam Electricity and mathematician with a large number of remarkable discoveries in classical geometry. Some of his results are shown and proven below.

Page made by vladimir.shelomovskii@gmail.com, vvsss

Quadrilateral with side bisectors

Dao ginma1.png
Dao ginma2.png

Let a convex quadrilateral $ABCD$ be given. Let $\ell', M'$ and $\ell, M$ be the bisector and the midpoint of $AB$ and $CD,$ respectively. Let $\ell'$ intersect $\ell$ at the point $P$ inside $ABCD.$ Denote $\angle APM' = \alpha, \angle MPD = \beta.$

Let point $S$ be the point inside $ABCD$ such that $\angle SDA = \alpha, \angle SAD = \beta.$

Let $Q$ be the point at ray $PM$ such that $\angle PDQ = \alpha.$ Define $Q'$ similarly.

1. Prove that $SQ \perp AB, SQ' \perp CD, SQ = PQ'.$

2. Prove that $\triangle BSC \sim \triangle ASD.$

Proof

1. $\angle SAD = \angle QPD, \angle SDA = \angle QDP \implies$ \[\triangle DAS \sim \triangle DPQ.\]

The spiral similarity taking $A$ to $S$ and $P$ to $Q$ has center $D$ and angle $\alpha.$ Therefore spiral similarity taking $AP$ to $SQ$ and $P$ to $Q$ has the same center $D$ and angle $\alpha.$

$\angle APM' = \alpha,$ so $AP$ maps into segment parallel $PM' \implies SQ \perp AB.$

2. Let $T$ be the spiral similarity centered at $D$ with angle $\alpha$ and coefficient $\frac {1}{|k|}, S = T(A), \frac {AD}{SD} = k.$

Let $t$ be spiral similarity centered at $C$ with angle $\alpha$ and coefficient $k, B = t(S), \frac {BC}{BS} = k.$

It is trivial that $t(T(P)) = P.$

It is known ( Superposition of two spiral similarities) that $B' = t(T(A))$ is the point with properties \[AP = BP, \angle APB = 2 \alpha \implies\] \[B' = B, \triangle BSC \sim \triangle ASD.\]

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Dao_Thanh_Oai_geometric_results&oldid=259143"