A Powerful Ratio Theorem

by math_pi_rate, Sep 27, 2018, 12:43 PM

So this topic is for a wonderful theorem about isogonal lines, namely Steiner's Ratio Theorem. Here's the theorem:

THEOREM (Steiner) Let $D$ and $E$ be points on $BC$, so that $AD$ and $AE$ are isogonal with respect to $\angle BAC$. Then we have the following equality:- $$\left (\frac{AB}{AC} \right)^2=\frac{BD \cdot BE}{CD \cdot CE}$$
REMARK: The converse of Steiner's Theorem is also true.

PROOF 1 (Trig Bash :D) Let $\angle BAD=\angle CAE=x$ and $\angle BAE=\angle CAD=y$. Apply sine law in $\triangle ADB, \triangle ADC, \triangle AEB, \triangle AEC$. Then we have $$\frac{BD}{AB}=\frac{\sin x}{\sin \angle BDA},\frac{BE}{AB}=\frac{\sin y}{\sin \angle BEA},\frac{CD}{AC}=\frac{\sin y}{\sin \angle CDA},\frac{CE}{AC}=\frac{\sin x}{\sin \angle CEA}$$Multiplying the first two fractions, dividing them by the next two fractions, and using the fact that $\sin(180^{\circ}-\theta)=\sin \theta$, one gets the required equality. The converse can be proved in a similar way.

PROOF 2 (Inversion) Let $\omega$ be the circumcircle of $\triangle ABC$, and let $AD \cap \omega=P, AE \cap \omega =Q$. Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the internal angle bisector of $\angle BAC$. Then $D \rightarrow Q$ and $E \rightarrow P$. Thus it suffices to show that $$\frac{BP}{CQ}=\frac{CP}{BQ} \Leftrightarrow \frac{BP}{CP}=\frac{CQ}{BQ}$$As $\angle BPC=\angle BQC$, this is equivalent to proving that $\triangle BPC \sim \triangle CQB$. But this is obviously true, cause we have $PQ \parallel BC$, i.e. $BPQC$ is an isosceles trapezoid. The converse can be proved in a similar fashion.

So now let's move on to some questions which actually use this theorem. Remember that this theorem is just a part of these solutions (albeit an important one), and so most solutions using this theorem might require other lemmas and observations too. Having said that, let's move forward.

PROBLEM 1 (Source=Balamatda)
Let $ABC$ be a triangle inscribed in a circle $(O)$. Suppose that $AH$ is the altitude and the line $AO$ intersects $BC$ at $D$. The circumcircle of $ADC$ intersects the circumcircle of $AHB$ at $E$. The tangent at $A$ of $(O)$ meets $BC$ at $I$. Prove that $IA=IE$.
https://scontent.fdad3-3.fna.fbcdn.net/v/t1.0-9/42614647_2256121971285643_3789226474465132544_n.jpg?_nc_cat=100&oh=dddf8d8f46971fe51fc3e49a6899879e&oe=5C61361D

SOLUTION: We have $\angle BEH=\angle BAH=\angle CAO=\angle CED \Rightarrow ED$ and $EH$ are isogonal wrt $\angle BEC$. By Steiner's Ratio Theorem, we get that $$\left (\frac{AB}{AC} \right)^2=\frac{BH \cdot BD}{CH \cdot CD}=\left (\frac{BE}{CE} \right)^2 \Rightarrow \frac{AB}{AC}=\frac{BE}{CE}$$This means that $E$ lies on the $A$-Apollonius circle, giving that $IA=IE$ (as $I$ is the center of the $A$-Apollonius circle). $\blacksquare$

REMARK: The above solution also shows that the problem is true for any two isogonal lines.


PROBLEM 2 (ELMO 2016 P6)
Elmo is now learning olympiad geometry. In triangle $ABC$ with $AB\neq AC$, let its incircle be tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$, respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$. Finally, let $\gamma$ be the circumcircle of $\triangle AST$.

(a) Help Elmo show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$.

(b) Help Elmo show that $\gamma$ is tangent to the incircle of $\triangle ABC$.

James Lin

SOLUTION: WLOG assume $AB \leq AC$. Let $I$ be the incenter of $\triangle ABC$.

Let $\omega$ be the circumcircle of $XSYT$, and $Z$ be the midpoint of $XY$, i.e. center of $\omega$. Also, Let $AI \cap BC = P$.

Let $DA$ meet the incircle again at $K$. Then $DEKF$ is harmonic $\Rightarrow -1=(K,D;E,F) \overset{D}{=} (A,P;X,Y)$.

This means that $A$ and $P$ are inverses w.r.t. $\omega$, i.e. $ZX^2=ZP \cdot ZA = PZ(PZ+PA) =PZ^2+PZ \cdot PA$

$\Rightarrow PZ \cdot PA = ZX^2-ZP^2 = (ZX-ZP)(ZX+ZP) = (ZX-ZP)(ZY+ZP) = PX \cdot PY = PS \cdot PT \Rightarrow ASZT$ is cyclic.

As $ZS = ZT$, $AZ$ is the internal angle bisector of $AS$ and $AT$ $\Rightarrow AS$ and $AT$ are isogonal.

(a) By Steiner's Ratio Theorem, $\frac{BT}{CT} \cdot \frac{BS}{CS} = \left(\frac{AB}{AC} \right)^2 \Rightarrow AC \cdot \sqrt{BS \cdot BT} - AB \cdot \sqrt{CS \cdot CT} = 0$. By the Converse of Casey's Theorem on point circles $\odot (A), \odot (B), \odot (C)$ and $\odot (AST)$, we get that $\odot (AST)$ and $\odot (ABC)$ are tangent to each other at $A$. $\blacksquare$

(b) Let $AI \cap EF = T$, then $\angle FTY = 90^{\circ} \Rightarrow \angle DYX = \angle FYT = 90^{\circ}-\angle EFD = \angle IDE = \angle IDX$

$\Rightarrow ID$ is tangent to $\odot (DXY) \Rightarrow ID^2 = IX \cdot IY \Rightarrow X$ and $Y$ are inverses w.r.t. the incircle.

Thus, $\omega$ and the incircle are orthogonal $\Rightarrow$ Length of tangent from $Z$ to $\omega$, i.e. $\ell$, is equal to $ZS$.

Now, $ZS \cdot ST = ZS(SD+TD) = ZS \cdot TD+ZT \cdot SD \Rightarrow ZS \cdot TD +ZT \cdot SD - ST \cdot \ell = 0$. By the Converse of Casey's Theorem on point circles $\odot (Z), \odot (S), \odot (T)$ and the incircle, and using the fact that $Z$ lies on $\odot (AST)$, we get that $\odot (AST)$ and the incircle are tangent to each other. $\blacksquare$

REMARK: The first part gives an important result, which has been stated more clearly below.


RESULT Let $D$ and $E$ be points on $BC$, so that $AD$ and $AE$ are isogonal with respect to $\angle BAC$. Then $\odot (ADE)$ is tangent to $\odot (ABC)$ at $A$. The converse is also true.

PROOF: See the solution to the first part of ELMO 2016 P6 given above.


PROBLEM 3 (Sharygin 2018 Correspondence Round Problem 16)
Let $ABC$ be a triangle with $AB < BC$. The bisector of angle $C$ meets the line parallel to $AC$ and passing through $B$, at point $P$. The tangent at $B$ to the circumcircle of $ABC$ meets this bisector at point $R$. Let $R'$ be the reflection of $R$ with respect to $AB$. Prove that $\angle R'PB = \angle RPA$.

SOLUTION: (Bary bash :D) The attached solution is the one that I submitted during the actual contest.
Attachments:
Solution_Q 16.docx (72kb)
This post has been edited 3 times. Last edited by math_pi_rate, Feb 13, 2019, 4:18 PM
Steiner
ratio
isogonal
reflection
Inversion
overlay

Comment

3 Comments

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Another question which uses this Theorem (as well as the result given after Problem 2) is IMOSL 2016 G2.

by math_pi_rate, Oct 12, 2018, 5:27 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
"Let $D$ and $E$ be points on $BC$, so that $AD$ and $AE$ are isogonal with respect to $\angle BAC$. Then $\odot (ADE)$ is tangent to $\odot (ABC)$ at $A$."

You can also do this by a homothety at $A.$ Or, as usual, use $\sqrt{bc}.$

by Wizard_32, Dec 15, 2018, 4:33 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Nice post! :wacko: I'm just getting confused....
shouldn't it be $\left (\frac{AB}{AC} \right)^2=\frac{BD \cdot BE}{CD \cdot CE}$ instead of $\left (\frac{AB}{AC} \right)^2=\frac{BD \cdot BE}{CD \cdot CA}$
Pls tell me if I'm wrong :wacko:

Sorry! Edited.
This post has been edited 1 time. Last edited by math_pi_rate, Feb 13, 2019, 4:27 PM

by AlastorMoody, Feb 12, 2019, 8:21 AM

Geo Geo everywhere, nor a point to see.

avatar

math_pi_rate
Shouts
Submit

  • First comment of January 26, 2025!

    by Yiyj1, Jan 27, 2025, 1:02 AM

  • And here's the first of 2025!

    by CrimsonBlade273, Jan 9, 2025, 6:16 AM

  • First comment of 2024!

    by mannshah1211, Jan 14, 2024, 3:01 PM

  • Wowowooo my man came backkkk

    by HoRI_DA_GRe8, Nov 11, 2023, 4:21 PM

  • Aah the thrill of coming back to a dead blog every year once!

    by math_pi_rate, Jul 28, 2023, 8:43 PM

  • kukuku 1st comment of 2023

    by kamatadu, Jan 3, 2023, 7:32 PM

  • 1st comment of 2022

    by HoRI_DA_GRe8, May 25, 2022, 8:29 PM

  • duh i found this masterpiece after the owner went inactive noooooooooooo :sadge: :(

    by Project_Donkey_into_M4, Nov 23, 2021, 2:56 PM

  • Hi everyone! Sorry but I doubt if I'll be reviving this anytime soon :oops:

    I might post something if I find anything interesting, but they'll mostly be short posts, unlike the previous ones!

    by math_pi_rate, Nov 7, 2020, 7:39 PM

  • @below be patient. He must be busy with some other work. :)

    by amar_04, Oct 22, 2020, 10:23 AM

  • Advanced is over now, please revive this please!

    by Geronimo_1501, Oct 12, 2020, 5:10 AM

  • REVIVE please

    by Gaussian_cyber, Aug 28, 2020, 2:19 PM

  • re-vi-ve

    by nprime06, Jun 12, 2020, 2:17 AM

  • re-vi-ve

    by fukano_2, May 30, 2020, 2:09 AM

  • Try this: Prove that a two colored cow has an odd number of Agis Phesis

    by Synthetic_Potato, Apr 18, 2020, 5:29 PM

68 shouts
Tags
geometry
Inversion
altitude
ex-point
humpty point
orthocenter
APMO
arc midpoint
Casey theorem
Circumcenter
collinearity
external angle bisector
Functional Equations
IMO
IMOSL 2002 G7
IMOSL 2011 G4
isogonal
length bash
overlay
perpendicular diagonals
Pointwise Value Trap
queue-point
ratio
reflection
Steiner
tangency of circles
About Owner
  • Posts: 1218
  • Joined: May 25, 2018
Blog Stats
  • Blog created: Aug 19, 2018
  • Total entries: 9
  • Total visits: 25927
  • Total comments: 30
Search Blog