# Brilliant Geometry contest- Season 1

Welcome to the first ever Brilliant Geometry contest.This contest aims to improve the brilliant user's ability to solve olympiad level Geometry problems.

This contest was originally held by ayush rai but he decided to hand the leadership over to me.

The rules of the contest are:

1. I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

2. A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

3. Please make a substantial comment.

4. Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 36 hours. Then, you must post the solution and you have the right to post a new problem.

5. If the one who solves the last problem does not post a new problem in 24 hours, the creator of the previous problem has the right to post another problem.

6. It is NOT compulsory to post original problems. But make sure it has not been posted on Brilliant.

7. If a diagram is involved in your problem please make sure it is drawn by a computer program.

8. Format your solution in $\LaTeX$, no picture solution will be accepted. Also make sure your solution is detailed and make sure to proof all claims.

SOLUTION TO PROBLEM n

proof here

Question as

PROBLEM n

Note by Aareyan Manzoor
4 years, 2 months ago

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Problem 4

Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$ (IMO Shortlisted)

- 4 years, 2 months ago

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, $AP' = AQ'$.

- 4 years, 2 months ago

Great solution. No doubt you were in IMO training!

- 4 years, 2 months ago

This works, tho I dont think you needed the extra effort to prove that P's dependancy, it would have been much easier to do separate cases for the proof since both cases are independant themselves.
A much clever way is to use directed angles to even skip the second case!

- 4 years, 2 months ago

True, but I'm used to making sure I'm not diagram dependent.

- 4 years, 2 months ago

PROBLEM 1

Let $X$ and $Y$ be points inside equilateral triangle $ABC$. Let $Y'$ be the reflection of $Y$ in line $BC$. Prove that

$XY+XB+XC \geq Y'A$

note: this problem was originally made by @Sharky Kesa in ayush's thread, i decided to start the contest with this, all credit goes to him

prince loomba posted his solution first but since his solution was wrong the credit goes to sharky kesa, he might post the next problem

- 4 years, 2 months ago

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

If $XY+XB+XC,

\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 $XY+XB+XC \geq Y'A$.

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

OK, you havent considered allcases. You specifically checked centroid, but what about an arbitrary point?

- 4 years, 2 months ago

I told my solution to KVPY 22nd ranker. He told my solution is absolutely correct

- 4 years, 2 months ago

The LHS cannot be minimum at just $X$ being the centroid, since $Y$ is still variable. This is where you have gone wrong.

- 4 years, 2 months ago

No. Y is A so lhs becomes XA+XB+XC which is min when X is centroid

- 4 years, 2 months ago

But you didn't state that in the statement! Thus, both $X$ and $Y$ have to be fixed for LHS to be minimum. Then, from here, you say that because of $Y$'s location, the RHS is maximised.

- 4 years, 2 months ago

Talk in slack. Here its difficult

- 4 years, 2 months ago

Thus, after careful discussion, this solution is indeed incorrect/incomplete.

- 4 years, 2 months ago

??? My solution is incorrect?

- 4 years, 2 months ago

PROBLEM 2

Let $D$, $E$ be points on sides $BC$, $AC$ respectively of triangle $ABC$ such that $DE||AB$. Let $X$ be a point outside the triangle such that segments $XD$ and $XE$ intersect side $AB$ at points $P$, $Q$ respectively. Find the smallest value $\gamma$ such that we must have

$|PQED| < \gamma |ABC|$

The problem has been solved by michael fuller

- 4 years, 2 months ago

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

- 4 years, 2 months ago

Neat solution! I'd have preferred not to use trig or co-ordinate geometry but this suffices as a solution. My solution is pretty similar but it goes like this:

Let $F$ and $G$ be the feet of the perpendiculars from $D$ and $E$ respectively on $AB$. Note that $FG$ must contain points $P$ and $Q$. We have

$|PQED| < |FGED| \leq \frac {1}{2} |ABC|$

This is sufficient to prove that $\gamma = \frac {1}{2}$.

- 4 years, 2 months ago

Great, I'll get to work on the next problem $:)$

- 4 years, 2 months ago

Neat Solution!

- 4 years, 2 months ago

Next problem?

- 4 years, 2 months ago

Problem 5

Let $D$, $E$, $F$ be points on side $BC$, $CA$, $AB$ respectively of triangle $ABC$. Prove that

$\min \{|AEF|, |BDF|, |CDE| \} \leq |DEF|$

- 4 years, 2 months ago

Here || represent?

- 4 years, 2 months ago

The area of that Triangle

- 4 years, 2 months ago

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!

- 4 years, 2 months ago

This proof is incorrect. Firstly, you have only considered two specific cases. Furthermore, for the rest of the cases, the area of $DEF$ gains, but so is true for the other three triangles, so you cannot apply this method.

- 4 years, 2 months ago

You dont understand my method! Read it carefully, especially the last paragraph!

- 4 years, 2 months ago

I do understand your proof. But it is incorrect. You say that the when the RHS is minimum, then LHS satisfies and when the LHS is minimum, the RHS satisfies. This doesn't imply that it is always satisfying. Furthermore, you give no proof of what the minimum value of the LHS is.

- 4 years, 2 months ago

Let f (x)>g (x).

If we prove that g (x) is less than the least value of f (x), and that f (x) is greater than the greatest value of g (x), dont you think this is enough for the proof of f (x)>g (x)?

- 4 years, 2 months ago

No it isn't. Example is $f(x)=\cos(x)$ and $g(x)=\sin(x)$. Say we claim that $f(x)>g(x)$. We have that the greatest value of $f(x)$ is $1$ at $x=2n\pi$, which is greater than $g(x)$. Also, the least value of $g(x)$ is $-1$ is at $x=\frac {3\pi}{2} + 2n\pi$, which is less than $f(x)$. Thus, by what you are saying, $f(x)>g(x)$, which is clearly incorrect.

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

Problem 6

Let $ABC$ be an isosceles triangle, with $\angle ABC = \angle ACB = 80^{\circ}$. Point $P$ is on $AB$ such that $AP = BC$. Find $\angle PCB$ with proof.

• Please refrain from using trigonometry. There are many Euclidean proofs available.

- 4 years, 2 months ago

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, $\angle BCP = 70^{\circ}$.

- 4 years, 2 months ago

What was the motivation for the construction?

- 4 years, 2 months ago

Another approach could be to observe that line joining C and P passes through the circumcenter of the triangle ABC.

- 4 years, 2 months ago

What was the motivation?

- 4 years, 2 months ago

The length $AP=BC$ is the motivation, since we want to try to relate these unrelated segments in some manner.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

If You have solved it You surely may!

And the answer is correct I guess

- 4 years, 2 months ago

Can you find a solution that doesn't use trigonometry? The problem is meant as a challenge to solve via only Euclidean geometry.

- 4 years, 2 months ago

But if i post a solution with trigonometry, will my solution be ignored?

- 4 years, 2 months ago

I would ignore it, since this contest is meant to be a challenge. However, you could share both trig and non-trig solution if you so wish to. This question is meant to help you think creatively for later, harder problems.

- 4 years, 2 months ago

Problem 7

$ABCD$ is an isosceles trapezium with $BC \parallel AD$ and $\angle CAD = 45^{\circ}$. $E$ is a point on $AB$ such that $DE \perp AB$. $F$ is a point on $DE$ such that $CF \perp DE$. Prove that $2 \angle FBD = \angle FCD$.

- 4 years, 2 months ago

Hey, This doesnot seem to be true.

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

- 4 years, 2 months ago

Sorry, I forgot one other condition.

- 4 years, 2 months ago

Ah, so now the problem becomes a simple angle-chase:

- 4 years, 2 months ago

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 $\angle FCD=\angle FGD=2\angle BFG$

- 4 years, 2 months ago

Problem 9

Let two circles $\Gamma_1$ and $\Gamma_2$ (with $\Gamma_1$ being larger than $\Gamma_2$) be internally tangent at $M$. $A$, $B$ and $C$ are points on $\Gamma_1$, such that $AB$ and $AC$ are tangent to $\Gamma_2$ at $P$ and $Q$ respectively. Let $MP$ and $MQ$ intersect $\Gamma_1$ for the second time at $X$ and $Y$ respectively. Prove that the intersection of $XC$ and $YB$ is on the line $PQ$.

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

Just as an extension, prove this intersection is the incentre of $\Delta ABC$.

- 4 years, 2 months ago

Follows from it lying on the angle-bisector and $AP=AQ$

- 4 years, 2 months ago

PROBLEM 10

Consider a right triangle $ABC$ with $m \angle C = 90°$. Let $D$ be the foot of the altitude from $C$. Let $E$ be a point in the on the line segment $CD$.

Let $F$ be the point on the segment $AE$ such that $BF = BC$. Similarly, let $G$ be the point on the segment $BE$ such that $AG = AC$. Let $X$ be the point of intersection of $AG$ and $BF$.

Show that $XF = XG$. (IMO Past Question)

- 4 years, 2 months ago

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 $XG=XF$.

- 4 years, 2 months ago

Beautiful solution,

Which year is this from btw?

- 4 years, 2 months ago

Thanks! I got no idea, but it was a really good question.

- 4 years, 2 months ago

Great job! I can reveal the problem was IMO 2012 Question 5.

- 4 years, 2 months ago

Ok That surprises me a bit, Aint this too easy for a P5!

No odffence to Sharky's great solution tho!

- 4 years, 2 months ago

PROBLEM 3

Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles - let $E$ be their intersection.

Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic, i.e. the four reflected points connect to form a cyclic quadrilateral.

This problem has been solved by Sualeh Asif.

- 4 years, 2 months ago

@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!

- 4 years, 2 months ago

Can you explain?

- 4 years, 2 months ago

Is this good?

- 4 years, 2 months ago

Yes, this works, I believe $:)$

My solution (or rather, the one I found with this problem) goes as follows:

Let $X$, $Y$, $Z$, $W$ be the feet of the altitudes from point $E$ to $\triangle AEB$, $\triangle BEC$, $\triangle CED$, $\triangle DEA$ respectively.

Note that the reflection of $E$ over the 4 lines is $XYZW$ with a scale factor of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections are also cyclic.

$\angle EWA$ is right angle and so is $\angle EXA$. Thus, $EXAW$ is cyclic with $EA$ being the diameter of the circumcircle.

It follows that $\angle EWX\cong\angle EAX\cong \angle EAB$ because they inscribe the same angle. Similarly $\angle EWZ\cong \angle EDC$, $\angle EYX\cong \angle EBA$, $\angle EYZ\cong \angle ECD$.

Futhermore, $m \angle XYZ+ m \angle XWZ \\ = m \angle EWX+ m \angle EYX+ m \angle EYZ+ m \angle EWZ \\ =360^{\circ}- m \angle CED- m \angle AEB \\ =180^{\circ}.$

Thus, $\angle XYZ$ and $\angle XWZ$ are supplementary and it follows that $XYZW$ is cyclic.

- 4 years, 2 months ago

Problem 8

Let $ABC$ be triangle with incenter $I$ . A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA = \angle PBC+\angle PCB$ . Show that $AP \geq AI$ , and that equality holds if and only if $P=I$.

(IMO shortlisted)

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

Problem 11

Definitions: Let $O$ be the centre of a circle $\Gamma$ with radius $r$. Let $K$ be a point. Let $K'$ be the point on $OK$ (possibly extended) satisfying $OK \times OK' = r^2$, with respect to $\Gamma$. Let $f(K)$ denote the line perpendicular to $K'$.

Given a cyclic quadrilateral $ABCD$. Let $AC \cap BD = P$, $AB \cap CD = Q$, $AD \cap BC = R$. Prove $f(P)=QR$, $f(Q)=PR$, $f(R)=PQ$.

- 4 years, 2 months ago

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 $PQR$

- 4 years, 2 months ago

But can you prove it?

- 4 years, 2 months ago

Surely, (I linked it above tho)

But Ill write the proof in a bit

- 4 years, 2 months ago

So here you go,(I assume a basic knowledge of projective geomtery)

Note that $f(Q)=PR\implies PR$is the polar of Q. So now we define a few more points:$PR\cap CD=E,PR\cap AB=F,AC\cap QR=G,BD\cap QR=H$, now just note that by perspecting through P, we have, $(Q,R;G,H)=(Q,E;C,D)=(Q,F;A,B)=-1$

Which simply implies that $EF=PR$ is the polar of $Q$. similarly $f(R)=PQ\implies PR$,

We can now end in several different ways, here is one of them that I like a lot.

1) Note that the two polar relationships imply that $O$ is the orthocenter of $PQR$, and hence $OP\perp QR$. But inverting in $(O)$, we have that the inverse of $P=AC\cap BD$ is the $M$, the miquel point of $ABCD$ , which we know by simple angle-chasing that it lies on $QR$. This combined with $OP\perp QR \implies P'=M \implies QR$ is the polar of $P \implies f(P)=QR$

- 4 years, 2 months ago

- 4 years, 2 months ago

Problem 12

Here is a problem I loved alot when I was given this, because of its utter simplicity:

Let the nine-point circle of $\triangle ABC$ intersect $BOC$ at $X,Y$.Prove that $\angle BAX=\angle CAY$

(I guess this is from Serbia)

- 4 years, 2 months ago

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 $O,H$

- 4 years, 2 months ago

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:

- 4 years, 2 months ago

1 week,12 problems.

- 4 years, 2 months ago

Problem 13:

$X$ and $Y$ are two points lying on or on the extensions of side $BC$ of $\triangle{ABC}$ such that $\widehat{XAY} = 90$. Let $H$ be the orthocenter of $\triangle{ABC}$. Take $X'$ and $Y'$ as the intersection points of $(BH,AX)$ and $(CH,AY)$ respectively. Prove that circumcircle $\triangle{CYY'}$,circumcircle of $\triangle{BXX'}$ and $X'Y'$ are concurrent.

Pakistan TST 2016 Problem 4, Iranian 3rd round Geometry exam P5 - 2014

- 4 years, 2 months ago