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
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