Brilliant Geometry contest- Season 1

Let the two intersection points of $EF$ with the circumcircle be $P$ and $P'$, with $P$ closer to $F$ than $E$.

Case 1: The question refers to point $P$.

Note that $BCEF$ and $ACDF$ are cyclic quadrilateral since the angles at $D$, $E$ and $F$ are right angles. We will now perform an angle chase. Let $\angle BAC = \alpha$, $\angle ABC = \beta$ and $\angle BCA = \gamma$. Thus, we have

$\begin{aligned} &\angle BDF = \angle FAC = \alpha\\ &\angle BFD = \angle ACD = \gamma\\ &\angle AFE = \angle BCE = \gamma \end{aligned}$

Since $\angle BPA = 180^{\circ} - \angle ACB = 180^{\circ} - \gamma$, $\angle PBA < \angle ACB = \gamma$. From this, we gather $\angle PBD + \angle BDF = \angle PBA + \angle ABD + \angle BDF < \alpha + \beta + \gamma = 180^{\circ}$. This implies that $Q$ is on the extensions of $BP$ and $DF$. This statement is required so as to prove I was not diagram dependent.

Because $APBC$ is a cyclic quadrilateral, $\angle QPA = \gamma$. But we also have $\angle AFE = \gamma$! Thus, $AQPF$ is cyclic. Since $\angle PFB = \angle AFE = \gamma$, we must have $\angle PQA = \gamma = \angle QPA$. Thus, $\Delta APQ$ is isosceles, so $AP = AQ$.

Case 2: The question refers to $P'$.

Let the intersection of $BP'$ and $DF$ be $Q'$. Using a similar argument as above, we prove I was not diagram dependent and that $Q'$ is on the interiors of $DF$ and $BP'$. We also have that since $AP'CB$ is cyclic, $\angle AP'Q' = \gamma$. We also have that $\angle BFD = \gamma$, so $P'AFQ'$ is cyclic. Since $\angle AFP'=\gamma$, we must have $\angle AQ'P' = \gamma$. Thus, $\Delta AP'Q'$ is isosceles. Therefore, $AP' = AQ'$.

Let $X'$ be the reflection of $X$ over $BC$.

If $XY+XB+XC<AY'$,

$\begin{aligned} & \implies X'Y' + X'B+X'C < AY'\\ & \implies X'B + X'C < AY' - X'Y'\\ & AY' - X'Y' \geq 0\\ & \implies X'B + X'C <0 \end{aligned}$

This is clearly untrue. Thus, we have a contradiction. Therefore $XY+XB+XC \geq Y'A$.

Solution to problem 1

We have to prove that LHS can be greater than RHS, let's take X=centroid of triangle than $XA=XB=XC$ and $Y$ be at the furthest point from $A$ which is $B$ and $C$. So $XY+XB+XC=3XA=3XB=3XC$. and $AY'=AY$. Clearly $3XA$ is greater than $YA$ because $Y$ is not close to being 3 times as far from $X$. SO, $XY + XB+XC >Y'A$.

Combining two equations we have the result

Proof for this to be the worst case

LHS is understood to be minimum when $X$ is centroid

RHS is maximum when $Y=A$. It can be proved by taking that $YY'$ is perpendicular to $BC$ and When $Y'$ is at maximum distance from $A$, this means that $YY'$ is maximum. Maximum value of $Y'$ can be $A$.

So by the rule of inequality, if we have to prove f (a)>f (b), we prove that the minimum value of f (a) is greater than the maximum value of f (b), So obviously the result is true.

??? My solution is incorrect?

I've got $\gamma=\dfrac{1}{2}$

Place the triangle base on the $x$ axis with $B$ at the origin and $A$ on the $x$ axis. Let the side length of the triangle be $s$. Let the line $y=a$ be between $C$ and the $x$ axis - it will cut the triangle at the two points $D$ and $E$ from left to right.

The point $X$ can be in the region below the line $AB$ and anywhere "trapped" inside the region bounded by the lines $BC$ extended and $AC$ extended. To find the smallest value of $\gamma$ we must find the largest value of $PQED$ and this is maximised as the $y$ ordinate of $X$ tends to $- \infty$, where $XD$ and $EQ$ are perpendicular to $AB$.

$|PQED|$ must be less than the maximum area of a rectangle inscribed in $\triangle ABC$. Using the line $y=a$, $|PQED| = a(s-2a \cot 60^{\circ}) = as-\dfrac{2 \sqrt3}{3}a^2$. Differentiating etc to find the optimal value of $a$ gives $a=\dfrac{\sqrt3}{4}s$ and thus $|PQED|=\dfrac{\sqrt3}{8}s=\dfrac{|ABC|}{2}$.

Therefore $\dfrac{|PQED|}{|ABC|}<\large \color{#20A900}{\boxed{\dfrac{1}{2}}}$.

Next problem?

Here || represent?

Here too I am using the same method.

First minimise area of $DEF$. This you will get when suppose D and F coincide with B and E with C. But here the area of triangle $FBD$ will become smaller. So LHS is less.

Then Maximise the minimum value of RHS. It can be 1/4 of the area of $ABC$ At max.

So here the equality holds.

After considering the worst cases, the equation is proved.

Proof

Let $f (x)\leq g (x)$ to be proved. If we prove that whenever f (x) attains its max value, then too g (x) is greater and when g (x) is minimum then too it is greater than f (x), our result is proved

Thats what I used here!

Solution

Here is my proof to this question, since no one has posted one themselves correctly.

Perform an affine transformation to convert $\Delta DEF$ into an equilateral triangle. Let $X$ be the point distinct from $D$ such that $\Delta XEF$ is also an equilateral triangle. WLOG $\angle BAC \geq 60^{\circ}$. Since $\angle BAC \geq \angle EXF$, $A$ must not exist outside the circumcircle of $XEF$. Thus, the area of $AEF$ is less than or equal to the area of $XEF$. But the area of $XEF$ is the same as the area of $DEF$. Thus, $|AEF| \leq |DEF|$. Therefore, $\min \{|AEF|, |BDF|, |CDE| \} \leq |DEF|$.

• An affine transformation is defined as the transformation of a plane in which you stretch or contract a plane in certain directions. Because of this, all triangles can be mapped into one another so it is always possible to map any arbitrary triangle into a specific one.

Here is one such Euclidean solution:

Construct point $X$ such that $X$ is closer $C$ than $B$ and $\Delta XAP \equiv ABC$. Angle chasing results in $\angle CAX = 60^{\circ}$. Notice that $\Delta ACX$ is isosceles, but $\angle CAX = 60^{\circ}$. Therefore, $\Delta ACX$ is equilateral. Further angle chasing results in $\angle PXC = 40^{\circ}$, and after a few angles, we get $\angle PCA = 10^{\circ}$. Therefore, $\angle BCP = 70^{\circ}$.

I used trigonometry and got the answer as 70 degrees. Should I post the solution?

Hey, This doesnot seem to be true.

However the following are true: $EA=EF$ and that $(EFA)$ is tangent to $CF$

Let $BD\cap AC=G$, Note that with our angle condition, $DFGC$ is cyclic, but also note that $\triangle CGD \sim \triangle BED \implies$ BD bisects $\angle CDF\implies$ $GFC$ is iscoceles, but also note that $GB=GC \implies G$ is the circumcenter of $\triangle BFC$

Finishing off we have $\angle FCD=\angle FGD=2\angle BFG$

Consider the hexagram $MXCABY$, then Pascal's Mystic Hexagram Theorem states that the intersections of the opposite sides: $MY$ and $AC$ ($Q$), $MX$ and $AB$ ($P$), and $BY$ and $CX$ (the intersection in question), are collinear.

This was an extremely tough question, I thought, but I think I've managed to scrape up a solution, only because a similar, simpler setup was considered during my selection camp for IMO. I would have no clue how to solve this without that. My solution will explore how to prove this through Power of a Point.

Firstly, we will create some more labels. Let the circle with radius $AG$ be $\Gamma_1$, and the circle with radius $BF$ be $\Gamma_2$. Also, Let $A'$ be the intersection of $AE$ and $\Gamma_2$, $B'$ be the intersection of $BE$ and $\Gamma_1$, and $C'$ be the reflection of $C$ over $AB$.

Some preliminary investigation leads us to conclude $AC$ is tangent to $\Gamma_2$ since $\angle ACB$ is $90^{\circ}$. Similarly, $BC$ is tangent to $\Gamma_1$.

We will now prove $F$, $G$, $A'$ and $B'$ are cyclic. Note that $C$ and $C'$ are the intersections of $\Gamma_1$ and $\Gamma_2$, so $CC'$ is the common chord of $\Gamma_1$ and $\Gamma_2$. We have

$EF \times EA' = EC \times EC' = EG \times EB'$

by Power of a Point. Thus, $EF \times EA' = EG \times EB'$, so $FGA'B'$ is a cyclic quadrilateral. Let the circle circumscribing this be $\Gamma_3$. We will now prove $BF$ is tangent to $\Gamma_3$. We have

$\begin{aligned} BF &= BC\\ BF^2 &= BC^2\\ &= BG \times BB' \end{aligned}$

The last statement is from Power of a Point in $\Gamma_1$. Now, from Power of a Point in $\Gamma_3$, since $BF^2 = BG \times BB'$, we have $BF$ is tangent to $\Gamma_3$. Similarly, $AG$ is tangent to $\Gamma_3$ as well. Thus, $XG$ and $XF$ are both tangents to $\Gamma_3$, so $XG=XF$.

@Michael Fuller this problem is trivial by inversion about E.

Let the Refleted points be $E_{ab},E_{bc},E_{cd},E_{da}$. Now consider the circles through the points $E,E_{ab},E_{ad}$ and so on. Call them $\Gamma_A,\Gamma_B,\Gamma_C,\Gamma_D$. Note that $\Gamma_A$ is centered at $A$ since $AE=AE_{ab}=AE_{ad}$. Now apply an inversion about E with arbitrary radius. Note that $\Gamma_A,\Gamma_C,$ are tangent and so are $\Gamma_B,\Gamma_D,$. Under the inversion $A',B',C',D'$ are arbitrary such that $A'C'\perp B'D'$. The critical observation is that $\Gamma_A '$ is inverted to a line perpendicular to $A'C'$ since $\Gamma_A$ is perpendicular to $AC$. Similarly we get two pairs of parallel lines $\Gamma_A ', \Gamma_C '$ and $\Gamma_B ', \Gamma_D '$. And hence the points $E_{ab} ',E_{bc} ',E_{cd} ',E_{da} '$ are just the vertices of a rectangle which is obviously cyclic. Hence $E_{ab},E_{bc},E_{cd},E_{da}$ are also cyclic!

Let $\angle A = \alpha$, $\angle B = \beta$ and $\angle C = \gamma$. We will first prove that $BIPC$ is a cyclic quadrilateral. We have:

$\angle PBA + \angle PCA +\angle PBC + \angle PCB= \beta + \gamma$, so $\angle PBC + \angle PCB = \frac{\beta+\gamma}{2}$. Note that this implies $\angle BPC = 90^{\circ} + \frac {\alpha}{2}$.

Since $I$ is the incentive of $ABC$, it can be angle chased to be shown that $\angle BIC = 180^{\circ} - \frac {\beta + \gamma}{2} = 90+\frac{\alpha}{2}$. Therefore, $\angle BPC = \angle BIC$, so $BIPC$ is a cyclic quadrilateral.

By Charles' Lemma, we must have the centre of this cyclic quad is the intersection of the line $AI$ extended with the circumcircle of $ABC$. Call this point $X$. We have the following:

$AP + PX \geq AX = AI + IX = AI + PX \implies AP \geq AI$

with equality if $I$ and $P$ concur.

Hey Sharky, this is a well-known Theorem-Brokard's Theorem:

It claims that $PQR$ is self-polar,(what you stated above), and that as a corollary $O$ is the orthocenter of $PQR$

Brokard

Let $M_b, M_c$ be the midpoints of $\overline{AC}, \overline{AB}$, respectively. Let $H$ be the projection of $A$ onto $BC$ and let $\ell$ be the internal bisector of $\angle BAC.$ Let $H_a$ be the foot of the A-perpendicular.

The transformation $\mathcal{T}$ composed of an inversion with center $A$ and power $\tfrac{1}{2} \cdot AB \cdot AC$ with a reflection in $\ell$ swaps $B, M_b$ and $C, M_c.$ Moreover, note that $\mathcal{T}(O) \in AH$ (because circumcenter and orthocenter are isogonal conjugates) and $B, C, \mathcal{T}(O)$ are collinear because $A, M_b, M_c, O$ are concyclic. Hence, $\mathcal{T}(O) \equiv H_a.$ It follows that $\mathcal{T}$ swaps $\odot(M_bH_a M_c)$ and $\odot(BOC).$ Then since $\odot(M_bH_aM_c)$ is just the nine-point circle of $\triangle ABC$, the intersections $X, Y$ of the nine-point circle and $\odot(BOC)$ are swapped by $\mathcal{T}.$ Hence, $AX, AY$ are isogonal WRT $\ell$, as desired.

(Solution by Dukejukem on AoPS.)

Challenge: There is a nice solution using similar triangles and angle-chasing, Find it.

Hint: Construct $O,H$

Here is the general problem and solution by Xuming Liang:

Let $C'\in AB,B'\in AC$ such that $B,C,B',C'$ are concyclic. Suppose $\omega, \omega'$ are two circles through $B,C$ and $C',B'$ respectively, such that they are corresponding circles between similar triangles $ABC, AB'C'$. If $\omega \cap \omega'=X,Y$, prove that $AX,AY$ are isogonals wrt $\angle BAC$.

Here are two solutions by him:

Serbian Problem