USA TST 2007, Problem 5 : Basic Geometry (Similar Triangles)

Question

I am given with a Triangle $ABC$ which is inscribed in circle $\omega$. The tangent lines to $\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS$ is perpendicular to $AT$. Points $B_1$ and $C_1$ lies on ray $ST$ (with $C_1$ in between $B_1$ and $S$) such that $B_{1}T=BT=C_{1}T$. I need to prove that the triangles $ABC$ and $AB_1C_1$ are similar to each other.

Answer 1

Here is a rewritten version of the OP with a picture. The order of "inventing" the points (in the solution using inversion) is the same as the order of the "inverted" points in a lemma. The lemma is simple, extracts the essence and gives the bridge for the solution. (Writing the lemma is not so simple, points have to be constructed in the right order.)

Proposition: There are given two circles with centers $O$ and $T$, which intersect orthogonally in two points, $B$ and $C$. The two circles are denoted by $(O)$, $(T)$, using their centers, if there is no danger of confusion.

Let $U=OT\cap BC$ be the mid point of the segment $BC$.

Let $S$ be a point on the ray = half-line $BC$, so that the circle with diameter $ST$ intersects $(O)$ in two points, $A$ and $a$. Here $A$ is on the bigger arc $\overset\frown{BC}$.

Let $B_1,C_1$ be the intersections of the line $ST$ with the circle $(T)$ , so that the points $B_1,T,C_1,S$ appear in this order on the line.

Then we have the following properties:

$(1)$ The points $B,T,C_1,A$ are on a circle.

$(2)$ The points $B,T,B_1,a$ are on a circle.

$(3)$ The points $A,B,B_1,S$ are on a circle.

$(1')$ The points $C,T,B_1,A$ are on a circle.

$(2')$ The points $C,T,C_1,a$ are on a circle.

$(3')$ The points $A,C,C_1,S$ are on a circle.

$(4)$ The triangles $\Delta ABB_1$, $\Delta ACC_1$ are similar.

$(5)$ The triangles $\Delta ABC$, $\Delta AB_1C_1$ are similar.

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Before we proceed, we consider the situation obtained by applying an inversion $X\to X'$ centered in $B$. Let $D$ be the intersection of $BT$ with the circle $(T)$, so that $BD$ is a diameter of $(T)$. By inversion, circles through $B$ are mapped into projective lines, so that $B\to B'=\infty$, and $\infty\to\infty'=B$. A line through $B$ is mapped in the same line. The inversion picture is then constructed using points from the following lemma.

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Lemma: Let $B$ be a point in the plane. Consider two perpendicular lines passing through $B$, and two points $A', T'$ on them respectively, so that we have $$BA'\perp BT'\ .$$

Let $D'$ be the mid point of segment $BT'$.

The perpendiculars in $D'$ and $A'$ on the lines $BD'T'$ and respectively $BA'$ intersect in a point $C'$.

Let $U'$ be the reflection of $B$ w.r.t. $C'$. (So $C'$ is the mid point of $BU'$.)

The circle $(A'U'T')$ intersects the lines $A'C'$ in $a'$, so that $A'a'$ is the diameter of this circle.

The line $A'T'$ intersects $C'D'$ in a point $B_1'$.

We consider the heights in triangle $\Delta A'a'B_1'$, and let its orthocenter be $C_1'$. Let $S'\in B_1'A'$ be the foot of the height from $a'$, $a'S'\perp B_1'A'$. and $BC'U'$ in $S'$.

The circle $(BS'T')$ intersects $D'C'$ in two points, $B_1', C_1'$, chosen so that $C_1'$ is between $D'$ and $C'$.

Then we have:

• $C_1'$ is the intersection of the diagonals in the parallelogram $A'D'T'C'$, so it is the mid point of $D'C'$, and of $A'T'$.
• The points $B, S', C_1', T', B_1'$ are on the circle.
• The points $B,S',C',U'$ are colinear.

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Proof of the lemma: Consider the four rectangles with a vertex in $C'$ which are congruent to the rectangle $A'C'D'B$, and the other vertices are among $A',B,D',T',U'$ or reflections of them w.r.t. $C'$. We see that $A'C'$ goes through the mid point of $T'U'$. So $A'C'$ is the side bisector of $T'U'$ (in the isosceles triangle $\Delta A'T'U'$), this implies that $A'a'$ is a diameter in the circumcircle $(A'S'T'a'U')$ of this triangle. So $A'T'\perp T'a'$.

By construction, $C_1'=A'T'\cap B_1'C'$ (intersection of two heights), so $C_1'= D'C'\cap A'T'$ is the intersection of the diagonals in the parallelogram $A'C'T'D'$.

The triangles $\Delta B_1'C_1'T'$, $\Delta B_1'C_1'S'$ have both a right angle opposite to the side $B_1'C_1'$, so $B_1'C_1'$ is the diameter of the circle $B_1'T'C_1'S'$, and on this circle there is also $B$, the reflection of $T'$ w.r.t. the same diameter.

We need now to show that $B',S',C'$ are colinear. For this we compute the angle $$ \begin{aligned} \widehat{BS'C'} &= \widehat{BS'B_1'} +\widehat{B_1'S'C_1'} +\widehat{C_1'S'C'} \\ &= \widehat{BT'B_1'} +90^\circ +\widehat{C_1'A'C'} \\ &= \widehat{A'a'T'} +90^\circ +\widehat{T'A'C'} \\ &= 90^\circ +90^\circ =180^\circ\ . \end{aligned} $$ This finishes the lemma.

$\square$

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Proof of the proposition: Lines of colinearity containing points $X',Y',Z',\dots$ in the lemma correspond to circles through $B$ containing points inverted points $X,Y,Z$. So we have the claimed propositions $(1)$, $(2)$, $(3)$; $(1')$, $(2')$, $(3')$. Let us use these circles to obtain equalities of angles. We have: $$ \begin{aligned} \widehat{ABS} &= \widehat{AB_1S} &&\text{ since $(ABB_1S)$ cyclic.}\\ \widehat{ACS} &= \widehat{AC_1S} &&\text{ since $(ACC_1S)$ cyclic. Passing to suplements:}\\ \widehat{ACB} &= \widehat{AC_1B_1}\ . \\ \widehat{CAC_1} &= \widehat{CSC_1} &&\text{ since $(ACC_1S)$ cyclic,}\\ &= \widehat{BSB_1} \\ &= \widehat{BAB_1} &&\text{ since $(ABB_1S)$ cyclic. This implies}\\ \widehat{BAC} &= \widehat{B_1AC_1} \ . \end{aligned} $$

$\square$

Answer 2

Probably a different solution than the previous answer, hence sharing it!

My Proof: Beautiful problem indeed!

Since $T$ is the point of intersection of tangent at $B$ and $C$ wrt $\omega$ , we get that $BT=CT$.

Now, using the conditions given us in the problem, we get that $CT=B_{1}T=C_{1}T$.

Hence $BCC_1B_1$ is cyclic quad.

Now, define $D$ as the midpoint of $BC$. Hence $TD\perp BC \implies TD \perp DS$ .

Since, $AS\perp AT$ ( given in the question ) , we get that $TDAS$ is cyclic.

Now, before proceeding further, we will state a lemma :

Lemma: Let $T$ be the intersection of the tangents to $(ABC)$ at $B$ and $C$ . Then line $AX$ is a symmedian , where a symmedian is the isogonal of a median in a triangle .

Proof: It can be directly proved using sin law!

Hence, by the above lemma , we note that $AT$ is the $A-$symmedian we have $\angle BAD=\angle TAC$.

Now, we are ready for angle chase!

Let $\angle BAD=\angle TAC= \theta$ , $\angle BAC= \alpha$ , $\angle ABC= B $ .

Claim : $ACC_1S$ is cyclic

Proof : Note that by tangent- side theorem , we get that $\alpha= \angle CBT= \angle BCT \implies \angle DTC= 90-\alpha$ .

Also, since $AS\perp AT$, we get that $\angle CAS= 90-\theta$ .

Also, since $TDAS$ is cyclic, we get $\angle DAT= \alpha -2\theta= \angle DST \implies \angle DTS = 90- (\alpha -2\theta) \implies \angle CTC_1=2\theta \implies \angle TC_1C=90-\theta \implies \angle CC_1S=90+\theta$ .

Hence we have $\angle CAS= 90-\theta$ and $\angle CC_1S=90+\theta$ . Hence $ACC_1S$ is cyclic.

Claim: $BB_1AS$ is cyclic

Proof: Note that $\angle BAS= 90+ \alpha - \theta \implies \angle ASB= 90-( B+\alpha - \theta) $.

By using the previous observation, that $\angle DST = \alpha -2\theta$ , we get that $\angle AST=\angle ASB_1= 90- (B+\theta)$ .

Again, using the previous observations that , $\angle BTC= 180-2\alpha$ and $\angle CTC_1=2\theta$ , we get that $ \angle BTB_1= 2 \alpha -2\theta \implies \angle B_1BT=90-(\alpha - \theta )$ .

Hence we get that $\angle B_1BA= 90+(B+\theta)$ .

Now since, $\angle ASB_1= 90- (B+\theta)$ and $\angle B_1BA= 90+(B+\theta)$ , we get that $BB_1AS$ is cyclic .

Now , the most beautiful part of this problem, spiral similarity !

Note that since $BB_1AS$ is cyclic and $ACC_1S$ is cyclic .Note that there is a spiral symmetry centered at $A$ dilating $\Delta ABC$ to $\Delta AB_1C_1$ .

Hence $\Delta ABC \sim \Delta AB_1C_1$. And we are done!

Note: Can someone please verify? I am very new to spiral similarity ( I just learned it today ).

Thanks in advance.