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The circle w is internally tangent with the circumcircle of $\triangle ABC$ at point A. $W \cap AB = K$ and $w$ intersects $BC$. The line $CL$ is tangent to $w$ at point L and $KL \cap BC = T$. Prove that $BT = BM$ where $BM$ is tangent to $w$ at point $M.$

What I tried:

$BM^2 = AB.BK$

I tried to prove that $\triangle$ ABT $\sim$ $\triangle$ TBK which means that $\frac {BT}{AB}$ = $\frac {BK}{BT}$ , but I couldn't.

I also couldn't prove that BM and BT are radiuses of the circle c with center point B.

Could somebody give me an idea?

enter image description here

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  • $\begingroup$ What is $W$? (in the second $K$ is defined as intersection of...) Please try to introduce first a letter, the to use it. It is for me irritating to read "The line $CL$ is..." because already in this second i am asking which $L$?! $\endgroup$ – dan_fulea Aug 8 '18 at 22:54
  • $\begingroup$ the small circle w, w ∩ AB = K $\endgroup$ – john1672 Aug 8 '18 at 22:57
  • $\begingroup$ That is a big $W$. And the one circle that appears in the picture and intersects $AB$ intersects it in two points, $A$ and $K$. Is it so complicated to Introduce things in order with full propositions? For instance: Let $ABC$ be a triangle. Let $(?)$ be the circumcircle of it. Let $(w)$ be a circle which is tangent to $(?)$ in $A$. The segment $AB$ intersects $(w)$ in $A$, and a second point $K$. It is really hard to digest the OP! $\endgroup$ – dan_fulea Aug 8 '18 at 23:03
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I would like to prove more, somehow "everything" in the given constellation. Also because of this, i will solve the "chronologically reversed problem", the one in which the appearance of the points is changed. Changing the order has (for me) the big advantage of constructing more points with easy to use angle properties, and thus with more "usable information".

The circle $(ABC)$ is "too heavy" to handle, so i considered $J=AC\cap (w)$, and imposed the condition $JK\| BC$ (equivalently) instead of the condition that the circles $(ABC)$ and $(w)$ touch each other in $A$. (So they can be obtained from each other by a homothety centered in $A$.)

For the start, let us show a simple lemma that makes clear which order of appearance is changed, the lemma shows the needed uniqueness property, that allows to change the order.

Lemma:

Let $(w)$ be a circle. Let $BC$ be a segment that intersects $(w)$ in two points. Let $\mathcal H$ be one of the two half-planes delimited by the line $BC$.

Then there exists a unique point $A\in\mathcal H$ on $(w)$ such that after constructing $K,J$ as the intersections $K = (AB)\cap(w)$, and $J=(AC)\cap (w)$, the line $KJ$ is parallel to $BC$.

(The statement is built so that every point is first introduced, than used. I was really missing this in the OP.)

Proof: Let $K^*$, $J^*$ be the touching points in $\mathcal H$ of tangents from $B$, respectively $C$ to the circle $(w)$. Let $K_0$, $J_0$ be the intersection points of $(w)$ with the segment $BC$, the order $B, K_0,J_0, C$ being respected.

Let $x^*$ be the minimal value among the measures of the arcs $K_0K^*$, and $J_0J^*$. Let $x$ be a parameter living in the interval $[0,x^*]$.

For each value of $x$ we have unique points $K=K(x)$, $J=J(x)$ on $(w)$, on the arcs $K_0K^*$, and $J_0J^*$ such that $${\color{red}x} =\overset\frown{K_0K} =\overset\frown{J_0J} \ . $$

Problem 2876649 Uniqueness of $A$

The point $A(x)$ is defined as the intersection of the two rays $BK(x)$ and $CJ(x)$. The function $x\to A(x)$ is continuous, so that the distance from $A(x)$ to $BC$ is strictly increasing. This strict monotone behavior shows the uniqueness. For $x\to 0$ the point $A(x)$ is in the interior of $(w)$. For the maximal value $x^*$, it is in the exterior. So there exists a (unique) point $x_w$ such that $A(x_w)$ is on the circle $(w)$.

$\square$

The following Proposition gives in fact the construction of $A$, reversing the creation of objects in the original post. By the proven uniqueness, we have the same constellation of points.

The posted problem follows from the following construction of $A$, and from the listed properties of the figure:

Proposition: Let the circle $(w)$ (with center $W$), the segment $BC$, and the half-plane $\mathcal H$ be as in the previous Lemma.

We consider the tangents from $B$, $C$ to $(w)$ in the half-plane $\mathcal H'$, opposite to $\mathcal H$. Let $S$ be their intersection. Let $M$, and respectively $L$ be their touching points with the circle.

Problem 2876649 MSE, enriched problem

Let $I$ be the in-center of the triangle $\Delta SBC$, the intersection of its angle bisectors. We draw these angle bisectors $BI$, $CI$, $SI$, and mark the delimited equal angles in $B$ with $y$, in $C$ with $z$, in $S$ with $s$. We denote by $\bar y$, $\bar z$, $\bar s$ the complement angles, i.e. $y+\bar y=z+\bar z=s+\bar s=90^\circ$.

We draw the line $ML$ and its intersection points with the bisectors $BI$, $CI$, denoted resp. by $X$ and $Y$.

Let $T\in BC$ be the symmetric point for $M$ w.r.t. $BI$.

Let $U\in BC$ be the symmetric point for $L$ w.r.t. $CI$.

The ray $(MU$ intersects the circle $(w)$ in a point named $J$.

The ray $(LT$ intersects the circle $(w)$ in a point named $K$.

Then we have:

  • (1) $IM=IL=IT=IU$, and the quadrilateral $MLUT$ is inscriptible.
  • (2) The line $KJ$ is parallel to $BC$.
  • (3) The lines $BK$ and $CJ$ intersect in a point $A$ on the circle $(w)$


    At this point the question in the OP is solved.


  • (4) The six points $A,C,L,X,I,T$ are on a circle.
  • (5) The six points $A,B,M,Y,I,U$ are on a circle.
  • (6) The lines $UL$ and $TM$ intersect in a point $\Xi$ on the circle $(w)$. Let $\Sigma$ be opposite w.r.t. the center $W$ on the circle. Then $\Xi$ and $\Sigma$ are the mid point of the two arcs $\overset \frown{JK}$, delimited by $J$ and $K$ on the circle. Moreover, $\Sigma$ is on $AI$.
  • (7) The lines $AI$, $JM$, $KL$ are concurrent.

Proof:

(1) From the symmetry w.r.t. the axes $BI$, $SI$, $CI$ for the pairs of symmetric points $(T,M)$, and $M,L$, and $L,U$, we obtain $IT=IM$, and $IM=IL$, and $IL=IU$, thus $I$ is the center of a circle $TMLU$ with radius $IT=IM=IL=IU$.

(2) We have $\angle UTL=\angle UML=\angle JML=\angle JKL$, so $JK\| UT$ (with equal correspondent angles built in $K,T$ w.r.t. line $KL$.

(3, 4, 5) We need to compute some angles, so that the position of $A$ is determined by a symmetical angle condition w.r.t. $y\leftrightarrow z$. The figure is determined (up to rescaling) by the angles $2y, 2z,2s$ in $\Delta BCS$, and by the position of $I$ on the angle bisector in $S$. Let us encode this position by the capture of the angle $$x=\widehat{IML}=\widehat{ILM}\ .$$

One more picture assisting the computations:

MSE problem 2876649 concurrence of AI, MJ, KL

We enumerate the following measures of angles.

The angles at the base $TM$ in $\Delta BTM$ are $\bar y$ and $\bar y$.

The angles at the base $UL$ in $\Delta CUL$ are $\bar z$ and $\bar z$.

The three angles in $L$, $\angle CLU$, $\angle ULM$, $\angle MLS$ have the measures respectively $\bar z$, $\bar y$, $\bar s$.

The three angles in $M$, $\angle BMT$, $\angle TML$, $\angle LMS$ have the measures respectively $\bar y$, $\bar z$, $\bar s$.

The angle $\angle ULI=\angle LUI$ is thus $\angle ULM-\angle ILM=\bar y-x=90^\circ-y-x$. This gives the angle at center in $I$ $$\widehat{LIU}=2(y+x)\ .$$ So the corresponding inscribed angle in $(TMLU)$ is $\widehat{LMU}=y+x$. We record this, the symmetric relation, and similar computations: $$ \begin{aligned} \frac 12\overset\frown{LJ} &= \widehat{LMJ}=\widehat{LMU}=\frac 12\widehat{LIU}=y+x\ ,\\ \frac 12 \overset\frown{MK} &= \widehat{MLK}=\widehat{MLT}=\frac 12\widehat{MIT}=z+x\ . \\ \frac 12\overset\frown{LM} &=\widehat {SML}=\bar s=y+z\ ,\\ \frac 12\overset\frown{KJ} &= 180^\circ-(2x+2y+2z)= 2(s-x) \ . \end{aligned} $$

Let now $A_1$ be the intersection of $CJ$ with the circle $(w)$. Then $\angle JA_1L=\angle JML=y+x$, and it may be better to display compactly this and the following information, isolating more angles realted to (inscribed angles pointing to) $\frac 12\overset\frown{JL}$: $$ \begin{aligned} \widehat{JA_1L} &=\frac 12\overset\frown{JL}=y+x\\ ,\\ \widehat{CTL} &=\widehat{JKL} = \frac 12\overset\frown{JL}=y+x\\ ,\\ \widehat{CIL} &=\frac 12\widehat{UIL} = \widehat{UML} = \widehat{JML} = \frac 12\overset\frown{JL}=y+x\\ . \end{aligned} $$ From here, the quadrilaterals $CA_1TL$, $CA_1IL$, $CTIL$ are inscriptible, so the five points $C,A,T,I,L$ are on a circle. It does not take much time to show, that $X$ is also on the same circle, since we have $$ \begin{aligned} \widehat{MXI} &= \widehat{MXB} = \widehat{MIB} - \widehat{IMX} \\ &= \frac 12\widehat{MIT} - x = \widehat{MLT} - x = \widehat{MLK} - x = (z+x)-x=z \\ &= \widehat{LCI}\ , \end{aligned} $$ so the interior angle in $C$ in $XLCI$ is equal to the exterior angle in the opposite vertex $I$, so $XLCI$ is inscriptible. So we have six points $$ C,\ A_1,\ T,\ I,\ X,\ L, $$ on a circle.

Let us now compute a "symmetric" (w.r.t. $x,y$) formula (in terms of $x,y,z$) for the angle $IA_1C$. We have for this $$ \begin{aligned} \widehat{IA_1C} &= \widehat{IA_1L} + \widehat{LA_1C} \\ &= \widehat{ICL} + \widehat{LA_1J} \\ &=z+(y+x)\\ &=y+z+x\ . \end{aligned} $$ We have computed $ \frac 12\overset\frown{KA_1J} = 180^\circ-2(y+z+x)$. So the angles $\angle CA_1I=\angle JA_1I$, and $\angle IA_1B=\angle IA_1K$ have the sum $2(x+y+z)$. This implies $\angle IA_1B=\angle IA_1K=(y+z)+x$, same value, symmetric expression w.r.t. the substitution $y\leftrightarrow z$.

An alternative argument uses the fact that $ATIC$ is inscriptible, and we have already computed its angle in $T$ against the chord $CI$, $y+z+x$.

This shows that $A_1$ is also on $BK$. To be precise, define $A_2$ as the intersection of $BK$ with the circle. The same computation leads to the same measure for the the arc length, $\overset\frown{ A_1K}=\overset\frown{ A_2K)$, so the two points $A_1$ and $A_2$ coincide.

From now on we use the notation $A$ for the point $A_1=A_2$.

By symmetry, the six points $$ B,\ A,\ U,\ I,\ Y,\ M, $$ are also on a circle.

This is a good point to insert one more picture:

MSE Problem 2876649 concurrence AI, MJ, KL, Pascal

Let us show now (6, 7). For (6), consider the angles $\angle CLJ$, and $\angle CLJ$. They are built between the tangent $CL$ and a chord of $(w)$, $C\Xi$, and respectively $CJ$. So the arc length of the corresponding arc is given by $$ \frac 12\overset\frown{J\Xi} = \angle CL\Xi-\angle CLJ = \bar z-(y+x)= 90^\circ-z-y-x\ . $$ This expression is symmetric in $y\leftrightarrow z$, "same computation" on the side of $\overset\frown{K\Xi}$ leads to the same result. This concludes (6).

We consider now $\Sigma$, the symmetric of $\Xi$ w.r.t. the center $W$ of $(w)$, so $\Xi\Sigma$ is a diameter.

Because $\Sigma$ is the mid point of "the other" arc $\overset\frown{JK}$, and $AI$ is the angle bisector of $\angle JAK$, this angle bisector passes through $\Sigma$.

The triangles $\Delta\Xi L\Sigma$ and $\Delta\Xi M\Sigma$ have a right angle in $L,M$, opposite to the diameter $\Xi\Sigma$.

Let $L',M'$ be on the circle $LMTU$, so that $L,\Sigma, L'$ on one side, and $M,\Sigma, M'$ on the other side, are colinear. Then we apply the theorem of Pascal in the circle $(LMTU)$ for the $6$-gon $$ LTM'MUL'\ , $$ so the following points are on a line: $$ LT\cap MU\ ,\qquad TM'\cap UL'= I\ ,\qquad M'M\cap L'L=\Sigma\ . $$ So the intersection $LT\cap MU$ lies on the line $AI\Sigma$, proving (7).

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  • $\begingroup$ what did you use to make the pictures?? $\endgroup$ – John C Aug 16 '18 at 7:47
  • $\begingroup$ geogebra on a linux box. It is available and free for both Linux and Win* operating systems. The solution was also found by (constructing points by intuition and) checking that some points are on a line / on a circle. The pictures can be easily deformed. Saving can be done in the formats ggb, geogebra specific, so that the "work" can be opened again in any geogebra session on an other machine, or as a picture, as above. $\endgroup$ – dan_fulea Aug 16 '18 at 7:55
  • $\begingroup$ I have upvoted this answer but, honestly, I'll need a whole day of my life just to read the solution from start to finish. I still doubt that this is the simplest possible solution. $\endgroup$ – Oldboy Aug 16 '18 at 15:36
  • $\begingroup$ This is not the simplest solution, but it is the simplest way to have all of the more or less obvious structure established. It is such a rare case to have a picture with two circles, each collecting six points of a given geometric constellation... If a simplest solution is wanted, i will extract it and type it... $\endgroup$ – dan_fulea Aug 16 '18 at 19:13
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Imagine that we move $C$ on fixed circle $c:=(ABC)$. Perform an inversion at $A$ (with arbitrary radius).

It takes circles $w$ and $c$ to parallel lines $w'$ and $c'$. Let $X'$ be a picture of $X$. Tangent $CL$ goes to circle $C'L'A$ which touches $w'$ at $L'$. Also lines $L-T-K$ and $C-T-B$ goes to circles $(L'T'K'A)$ and $(C'T'B'A)$ respectively.

enter image description here

Since $L'K'|| B'C'$ and $$\angle B'C'T' = \angle B'AT' = \angle K'L'T'$$ we see that $C',L'$ and $T'$ are collinear.

Now, since (tangent-chord property) $$\angle T'C'A = \angle AL'K' = \angle AT'K'$$ we see that $K'T'$ is tangent line on circle $(AB'C'T')$ in $T'$. So by PoP with respect to circle $(AB'C'T)$ and a point $K'$ we have $$K'T'^2 = K'A\cdot KB' = const. $$ so (while we move $C$) $T'$ is on fixed circle $\mathcal{C}$ with center at $K'$. Since $K'A \ne K'B'$ this circle $\mathcal{C}$ doesn't goes through $A$ so $K$ is on fixed circle $\mathcal{C}'$. Now it is not difficult to see $MT = MK$.

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