Fontene Theorems Elementary Proofs

by XmL, Aug 12, 2014, 10:26 PM

In this blog we will prove all three of the Fontene theorems in an elementary fashion with a different point of view. Either you want to know what they are beforehand or try proving them yourself they are here: http://mathworld.wolfram.com/FonteneTheorems.html

Let's begin by defining an important point:

Let $l$ be an arbitrary line that passes through the circumcenter of $\triangle ABC$, define $l\cap (O)=Y,Z$. The Simson lines of $Y,Z$ wrt $ABC$ meet at $P$.

Result: If $X\in l$ and $X\ne Y,Z$. Then the pedal circle of $X$ wrt $ABC$ passes through $P$.

Before reading on

Diagram

Proof: Let the pedal triangle of $P$ be $DEF$, The $Y,Z$ Simson lines $\cap BC=Y_A,Z_A$, $Y_B,Y_C,Z_B,Z_C$ are similarly defined.

Since $ZY$ is the diameter, we know that the Simson lines are orthogonal. From this we can discover a lot of similar triangles in the diagram(that can be proven by simple angle chasings). In particular, $PZ_BY_B\sim BZY, PZ_CY_C\sim CZY$. Note that since $ZZ_B\parallel XE\parallel YY_B$, therefore $\frac {Z_BE}{EY_B}=\frac {ZX}{YX}$, hence $E,X$ are corresponding points in the two similar triangles $\Rightarrow \angle EPY_B=\angle XBY$, symmetrically we have $\angle YPZ_C=\angle XCZ$. Hence $\angle YPE=90+\angle XBY+\angle XCZ=$ $90+(90-\angle ZBX)+(90-\angle YCX)=180-\angle ABX-\angle XCA$ $=180-\angle FDE$ $\Rightarrow P,E,D,F$ are concyclic which means $P\in (DEF)$ and we are done. $\Box$.

Correlary-2nd Fontene theorem:
If a point moves on a fixed line through the circumcenter, then its pedal circle passes through a fixed point on the nine-point circle.

Proof: Clearly $P$ defined above is the fixed point, moreover since the nine-point circle is the pedal circle of the circumcenter therefore $P$ lies on it. $\Box$.

1st Fontene theorem(Rephrased): The $A-$ midline,$NM,DP$ concur.

Proof: Let $M,N$ be the midpoints of $AC,AB$, $NM\cap FE=W$. Since $\frac {Z_BM}{Y_BM}=\frac {ZO}{YP}=1$, therefore $M$ is the midpoint of $Z_BY_B$, likewise $N$ is the midpoint of $Y_CZ_C$. Now construct $P'=(M,MP)\cap (N,NP)$, which is the miquel point of complete quad. $ANWEMF$ and the reflection of $P$ over $NM$. Hence we have $\angle FP'E=\angle NP'M=\angle NPM=180-\angle A=\angle FXE$ $\Rightarrow P'\in (AEFX)$

Next we will prove that $P'\in l$: It's easy to show that $P'Y_B\parallel CZ,P'Z_C\parallel BY, CYY_B\sim ZBZ_C$. Let $P'Y_B\cap CY,ZY=P_Y,U_Y, P'Z_C\cap U_Z,P_Y$. Since $U_Y,U_Z$ are two corresponding points of the similar triangles mentions, therefore $\frac {YP_Y}{ZY}=\frac {YU_Y}{CY}=\frac {BU_Z}{ZB}=\frac {YP_Z}{ZY}$ and hence $P_Y\equiv P_Z\equiv P'$.

Note that $AZY\sim PZ_AY_A$ and that $X,D$ are two corresponding points, therefore $\angle PDC=\angle AXP'=\angle MEP'=\angle MWP'=\angle PWM$, since $WM\parallel DC$, therefore $P,W,D$ are collinear.(Let $PD\cap NM=W'\Rightarrow \angle PW'M=\angle PWM\Rightarrow W\equiv W'$), and we are done. $\Box$.

Note: $l$ is the dilation of $P$'s Simson line wrt the medial triangle by a factor of $2$.

*To be continued*
This post has been edited 7 times. Last edited by XmL, Aug 12, 2014, 11:54 PM

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Still an immature proof writer :)

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  • Good blog! I hope to see more posts here!

    by WizardMath, Feb 3, 2018, 6:57 PM

  • Thanks Harry Potter :D

    by XmL, Jul 30, 2014, 10:47 PM

  • Hey dude. Nice blog. I like your work.

    by gamjawon, Jul 30, 2014, 7:43 AM

  • Thanks...although most of the posts here won't be about geometry...rightnow I'm trying to improve my skills in other areas such as number theory and combo.

    by XmL, Jun 5, 2014, 1:11 AM

  • Hi Geometer,,nice blog..:D

    by vanu1996, Jun 4, 2014, 1:27 PM

  • Check your private massage :)

    by zabihpourmahdi, Jan 11, 2013, 5:34 PM

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