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

Format your proofs as

SOLUTION TO PROBLEM n

proof here

Question as

PROBLEM n

ask question relevant question here

To answer the latest question just shift to the "newest" mode

Note by Aareyan Manzoor
3 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Problem 4

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

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Let the two intersection points of EFEF with the circumcircle be PP and PP', with PP closer to FF than EE.

Case 1: The question refers to point PP.

Note that BCEFBCEF and ACDFACDF are cyclic quadrilateral since the angles at DD, EE and FF are right angles. We will now perform an angle chase. Let BAC=α\angle BAC = \alpha, ABC=β\angle ABC = \beta and BCA=γ\angle BCA = \gamma. Thus, we have

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

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

Because APBCAPBC is a cyclic quadrilateral, QPA=γ\angle QPA = \gamma. But we also have AFE=γ\angle AFE = \gamma! Thus, AQPFAQPF is cyclic. Since PFB=AFE=γ\angle PFB = \angle AFE = \gamma, we must have PQA=γ=QPA\angle PQA = \gamma = \angle QPA. Thus, ΔAPQ\Delta APQ is isosceles, so AP=AQAP = AQ.

Case 2: The question refers to PP'.

Let the intersection of BPBP' and DFDF be QQ'. Using a similar argument as above, we prove I was not diagram dependent and that QQ' is on the interiors of DFDF and BPBP'. We also have that since APCBAP'CB is cyclic, APQ=γ\angle AP'Q' = \gamma. We also have that BFD=γ\angle BFD = \gamma, so PAFQP'AFQ' is cyclic. Since AFP=γ\angle AFP'=\gamma, we must have AQP=γ\angle AQ'P' = \gamma. Thus, ΔAPQ\Delta AP'Q' is isosceles. Therefore, AP=AQAP' = AQ'.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Great solution. No doubt you were in IMO training!

Prince Loomba - 3 years, 3 months ago

Log in to reply

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!

Sualeh Asif - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

PROBLEM 1

Let XX and YY be points inside equilateral triangle ABCABC. Let YY' be the reflection of YY in line BCBC. Prove that

XY+XB+XCYAXY+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

Aareyan Manzoor - 3 years, 3 months ago

Log in to reply

Let XX' be the reflection of XX over BCBC.

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

    XY+XB+XC<AY    XB+XC<AYXYAYXY0    XB+XC<0\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+XCYAXY+XB+XC \geq Y'A.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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=XCXA=XB=XC and YY be at the furthest point from AA which is BB and CC. So XY+XB+XC=3XA=3XB=3XCXY+XB+XC=3XA=3XB=3XC. and AY=AYAY'=AY. Clearly 3XA3XA is greater than YAYA because YY is not close to being 3 times as far from XX. SO, XY+XB+XC>YAXY + 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 XX is centroid

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

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.

Prince Loomba - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

Prince Loomba - 3 years, 3 months ago

Log in to reply

@Prince Loomba The LHS cannot be minimum at just XX being the centroid, since YY is still variable. This is where you have gone wrong.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

Prince Loomba - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

@Sharky Kesa Talk in slack. Here its difficult

Prince Loomba - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

??? My solution is incorrect?

Prince Loomba - 3 years, 3 months ago

Log in to reply

PROBLEM 2

Let DD, EE be points on sides BCBC, ACAC respectively of triangle ABCABC such that DEABDE||AB. Let XX be a point outside the triangle such that segments XDXD and XEXE intersect side ABAB at points PP, QQ respectively. Find the smallest value γ\gamma such that we must have

PQED<γABC|PQED| < \gamma |ABC|

The problem has been solved by michael fuller

Sharky Kesa - 3 years, 3 months ago

Log in to reply

I've got γ=12\gamma=\dfrac{1}{2}

Place the triangle base on the xx axis with BB at the origin and AA on the xx axis. Let the side length of the triangle be ss. Let the line y=ay=a be between CC and the xx axis - it will cut the triangle at the two points DD and EE from left to right.

The point XX can be in the region below the line ABAB and anywhere "trapped" inside the region bounded by the lines BCBC extended and ACAC extended. To find the smallest value of γ\gamma we must find the largest value of PQEDPQED and this is maximised as the yy ordinate of XX tends to - \infty, where XDXD and EQEQ are perpendicular to ABAB.

PQED|PQED| must be less than the maximum area of a rectangle inscribed in ABC\triangle ABC. Using the line y=ay=a, PQED=a(s2acot60)=as233a2 |PQED| = a(s-2a \cot 60^{\circ}) = as-\dfrac{2 \sqrt3}{3}a^2. Differentiating etc to find the optimal value of aa gives a=34sa=\dfrac{\sqrt3}{4}s and thus PQED=38s=ABC2|PQED|=\dfrac{\sqrt3}{8}s=\dfrac{|ABC|}{2}.

Therefore PQEDABC<12\dfrac{|PQED|}{|ABC|}<\large \color{#20A900}{\boxed{\dfrac{1}{2}}}.

Michael Fuller - 3 years, 3 months ago

Log in to reply

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 FF and GG be the feet of the perpendiculars from DD and EE respectively on ABAB. Note that FGFG must contain points PP and QQ. We have

PQED<FGED12ABC|PQED| < |FGED| \leq \frac {1}{2} |ABC|

This is sufficient to prove that γ=12\gamma = \frac {1}{2}.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

@Sharky Kesa Great, I'll get to work on the next problem :):)

Michael Fuller - 3 years, 3 months ago

Log in to reply

Neat Solution!

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Next problem?

Armain Labeeb - 3 years, 3 months ago

Log in to reply

Problem 5

Let DD, EE, FF be points on side BCBC, CACA, ABAB respectively of triangle ABCABC. Prove that

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Here || represent?

Prince Loomba - 3 years, 3 months ago

Log in to reply

The area of that Triangle

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Here too I am using the same method.

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

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

So here the equality holds.

After considering the worst cases, the equation is proved.

Proof

Let f(x)g(x)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!

Prince Loomba - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

Prince Loomba - 3 years, 3 months ago

Log in to reply

@Prince Loomba 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.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

@Sharky Kesa 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)?

Prince Loomba - 3 years, 3 months ago

Log in to reply

@Prince Loomba No it isn't. Example is f(x)=cos(x)f(x)=\cos(x) and g(x)=sin(x)g(x)=\sin(x). Say we claim that f(x)>g(x)f(x)>g(x). We have that the greatest value of f(x)f(x) is 11 at x=2nπx=2n\pi, which is greater than g(x)g(x). Also, the least value of g(x)g(x) is 1-1 is at x=3π2+2nπx=\frac {3\pi}{2} + 2n\pi, which is less than f(x)f(x). Thus, by what you are saying, f(x)>g(x)f(x)>g(x), which is clearly incorrect.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Solution

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

Perform an affine transformation to convert ΔDEF\Delta DEF into an equilateral triangle. Let XX be the point distinct from DD such that ΔXEF\Delta XEF is also an equilateral triangle. WLOG BAC60\angle BAC \geq 60^{\circ}. Since BACEXF\angle BAC \geq \angle EXF, AA must not exist outside the circumcircle of XEFXEF. Thus, the area of AEFAEF is less than or equal to the area of XEFXEF. But the area of XEFXEF is the same as the area of DEFDEF. Thus, AEFDEF|AEF| \leq |DEF|. Therefore, min{AEF,BDF,CDE}DEF\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.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Problem 6

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


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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Here is one such Euclidean solution:

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

What was the motivation for the construction?

Harsh Shrivastava - 3 years, 3 months ago

Log in to reply

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

Harsh Shrivastava - 3 years, 3 months ago

Log in to reply

@Harsh Shrivastava What was the motivation?

Harsh Shrivastava - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

A Former Brilliant Member - 3 years, 3 months ago

Log in to reply

If You have solved it You surely may!

And the answer is correct I guess

Sualeh Asif - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

Armain Labeeb - 3 years, 3 months ago

Log in to reply

@Armain Labeeb 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.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Problem 7

ABCDABCD is an isosceles trapezium with BCADBC \parallel AD and CAD=45\angle CAD = 45^{\circ}. EE is a point on ABAB such that DEABDE \perp AB. FF is a point on DEDE such that CFDECF \perp DE. Prove that 2FBD=FCD2 \angle FBD = \angle FCD.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Hey, This doesnot seem to be true.

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

Zoha Asif - 3 years, 3 months ago

Log in to reply

Sorry, I forgot one other condition.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Let BDAC=GBD\cap AC=G, Note that with our angle condition, DFGCDFGC is cyclic, but also note that CGDBED    \triangle CGD \sim \triangle BED \implies BD bisects CDF    \angle CDF\implies GFCGFC is iscoceles, but also note that GB=GC    GGB=GC \implies G is the circumcenter of BFC\triangle BFC

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

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Problem 9

Let two circles Γ1\Gamma_1 and Γ2\Gamma_2 (with Γ1\Gamma_1 being larger than Γ2\Gamma_2) be internally tangent at MM. AA, BB and CC are points on Γ1\Gamma_1, such that ABAB and ACAC are tangent to Γ2\Gamma_2 at PP and QQ respectively. Let MPMP and MQMQ intersect Γ1\Gamma_1 for the second time at XX and YY respectively. Prove that the intersection of XCXC and YBYB is on the line PQPQ.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Consider the hexagram MXCABYMXCABY, then Pascal's Mystic Hexagram Theorem states that the intersections of the opposite sides: MYMY and ACAC (QQ), MXMX and ABAB (PP), and BYBY and CXCX (the intersection in question), are collinear.

Michael Fuller - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

@Sharky Kesa Follows from it lying on the angle-bisector and AP=AQAP=AQ

Sualeh Asif - 3 years, 3 months ago

Log in to reply

PROBLEM 10

Consider a right triangle ABCABC with mC=90°m \angle C = 90°. Let DD be the foot of the altitude from CC. Let EE be a point in the on the line segment CDCD.

Let FF be the point on the segment AEAE such that BF=BCBF = BC. Similarly, let GG be the point on the segment BEBE such that AG=ACAG = AC. Let XX be the point of intersection of AGAG and BFBF.

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

Michael Fuller - 3 years, 3 months ago

Log in to reply

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 AGAG be Γ1\Gamma_1, and the circle with radius BFBF be Γ2\Gamma_2. Also, Let AA' be the intersection of AEAE and Γ2\Gamma_2, BB' be the intersection of BEBE and Γ1\Gamma_1, and CC' be the reflection of CC over ABAB.

Some preliminary investigation leads us to conclude ACAC is tangent to Γ2\Gamma_2 since ACB\angle ACB is 9090^{\circ}. Similarly, BCBC is tangent to Γ1\Gamma_1.

We will now prove FF, GG, AA' and BB' are cyclic. Note that CC and CC' are the intersections of Γ1\Gamma_1 and Γ2\Gamma_2, so CCCC' is the common chord of Γ1\Gamma_1 and Γ2\Gamma_2. We have

EF×EA=EC×EC=EG×EBEF \times EA' = EC \times EC' = EG \times EB'

by Power of a Point. Thus, EF×EA=EG×EBEF \times EA' = EG \times EB', so FGABFGA'B' is a cyclic quadrilateral. Let the circle circumscribing this be Γ3\Gamma_3. We will now prove BFBF is tangent to Γ3\Gamma_3. We have

BF=BCBF2=BC2=BG×BB\begin{aligned} BF &= BC\\ BF^2 &= BC^2\\ &= BG \times BB' \end{aligned}

The last statement is from Power of a Point in Γ1\Gamma_1. Now, from Power of a Point in Γ3\Gamma_3, since BF2=BG×BBBF^2 = BG \times BB', we have BFBF is tangent to Γ3\Gamma_3. Similarly, AGAG is tangent to Γ3\Gamma_3 as well. Thus, XGXG and XFXF are both tangents to Γ3\Gamma_3, so XG=XFXG=XF.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Beautiful solution,

Which year is this from btw?

Sualeh Asif - 3 years, 3 months ago

Log in to reply

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

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

Michael Fuller - 3 years, 3 months ago

Log in to reply

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

No odffence to Sharky's great solution tho!

Sualeh Asif - 3 years, 3 months ago

Log in to reply

PROBLEM 3

Let ABCDABCD be a convex quadrilateral such that diagonals ACAC and BDBD intersect at right angles - let EE be their intersection.

Prove that the reflections of EE across ABAB, BCBC, CDCD, DADA are concyclic, i.e. the four reflected points connect to form a cyclic quadrilateral.

This problem has been solved by Sualeh Asif.

Michael Fuller - 3 years, 3 months ago

Log in to reply

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

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

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Can you explain?

Michael Fuller - 3 years, 3 months ago

Log in to reply

@Michael Fuller Is this good?

Sualeh Asif - 3 years, 3 months ago

Log in to reply

@Sualeh Asif Yes, this works, I believe :):)

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

Let XX, YY, ZZ, WW be the feet of the altitudes from point EE to AEB\triangle AEB, BEC\triangle BEC, CED\triangle CED, DEA\triangle DEA respectively.

Note that the reflection of EE over the 4 lines is XYZWXYZW with a scale factor of 22 with center EE. Thus, if XYZWXYZW is cyclic, then the reflections are also cyclic.

EWA\angle EWA is right angle and so is EXA\angle EXA. Thus, EXAWEXAW is cyclic with EAEA being the diameter of the circumcircle.

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

Futhermore, mXYZ+mXWZ=mEWX+mEYX+mEYZ+mEWZ=360mCEDmAEB=180. 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, XYZ\angle XYZ and XWZ\angle XWZ are supplementary and it follows that XYZWXYZW is cyclic.

Michael Fuller - 3 years, 3 months ago

Log in to reply

Problem 8

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

(IMO shortlisted)

Sualeh Asif - 3 years, 3 months ago

Log in to reply

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

PBA+PCA+PBC+PCB=β+γ\angle PBA + \angle PCA +\angle PBC + \angle PCB= \beta + \gamma, so PBC+PCB=β+γ2\angle PBC + \angle PCB = \frac{\beta+\gamma}{2}. Note that this implies BPC=90+α2\angle BPC = 90^{\circ} + \frac {\alpha}{2}.

Since II is the incentive of ABCABC, it can be angle chased to be shown that BIC=180β+γ2=90+α2\angle BIC = 180^{\circ} - \frac {\beta + \gamma}{2} = 90+\frac{\alpha}{2}. Therefore, BPC=BIC\angle BPC = \angle BIC, so BIPCBIPC is a cyclic quadrilateral.

By Charles' Lemma, we must have the centre of this cyclic quad is the intersection of the line AIAI extended with the circumcircle of ABCABC. Call this point XX. We have the following:

AP+PXAX=AI+IX=AI+PX    APAIAP + PX \geq AX = AI + IX = AI + PX \implies AP \geq AI

with equality if II and PP concur.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

Problem 11

Definitions: Let OO be the centre of a circle Γ\Gamma with radius rr. Let KK be a point. Let KK' be the point on OKOK (possibly extended) satisfying OK×OK=r2OK \times OK' = r^2, with respect to Γ\Gamma. Let f(K)f(K) denote the line perpendicular to KK'.

Given a cyclic quadrilateral ABCDABCD. Let ACBD=PAC \cap BD = P, ABCD=QAB \cap CD = Q, ADBC=RAD \cap BC = R. Prove f(P)=QRf(P)=QR, f(Q)=PRf(Q)=PR, f(R)=PQf(R)=PQ.

Sharky Kesa - 3 years, 3 months ago

Log in to reply

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

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

Sualeh Asif - 3 years, 3 months ago

Log in to reply

But can you prove it?

Sharky Kesa - 3 years, 3 months ago

Log in to reply

@Sharky Kesa Surely, (I linked it above tho)

But Ill write the proof in a bit

Sualeh Asif - 3 years, 3 months ago

Log in to reply

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

Note that f(Q)=PR    PRf(Q)=PR\implies PRis the polar of Q. So now we define a few more points:PRCD=E,PRAB=F,ACQR=G,BDQR=HPR\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(Q,R;G,H)=(Q,E;C,D)=(Q,F;A,B)=-1

Which simply implies that EF=PREF=PR is the polar of QQ. similarly f(R)=PQ    PRf(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 OO is the orthocenter of PQRPQR, and hence OPQROP\perp QR. But inverting in (O)(O), we have that the inverse of P=ACBDP=AC\cap BD is the MM, the miquel point of ABCDABCD , which we know by simple angle-chasing that it lies on QRQR. This combined with OPQR    P=M    QROP\perp QR \implies P'=M \implies QR is the polar of P    f(P)=QRP \implies f(P)=QR

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Brokard

Sualeh Asif - 3 years, 3 months ago

Log in to reply

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 ABC\triangle ABC intersect BOCBOC at X,YX,Y.Prove that BAX=CAY\angle BAX=\angle CAY

(I guess this is from Serbia)

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Let Mb,McM_b, M_c be the midpoints of AC,AB\overline{AC}, \overline{AB}, respectively. Let HH be the projection of AA onto BCBC and let \ell be the internal bisector of BAC.\angle BAC. Let HaH_a be the foot of the A-perpendicular.

The transformation T\mathcal{T} composed of an inversion with center AA and power 12ABAC\tfrac{1}{2} \cdot AB \cdot AC with a reflection in \ell swaps B,MbB, M_b and C,Mc.C, M_c. Moreover, note that T(O)AH\mathcal{T}(O) \in AH (because circumcenter and orthocenter are isogonal conjugates) and B,C,T(O)B, C, \mathcal{T}(O) are collinear because A,Mb,Mc,OA, M_b, M_c, O are concyclic. Hence, T(O)Ha.\mathcal{T}(O) \equiv H_a. It follows that T\mathcal{T} swaps (MbHaMc)\odot(M_bH_a M_c) and (BOC).\odot(BOC). Then since (MbHaMc)\odot(M_bH_aM_c) is just the nine-point circle of ABC\triangle ABC, the intersections X,YX, Y of the nine-point circle and (BOC)\odot(BOC) are swapped by T.\mathcal{T}. Hence, AX,AYAX, 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,HO,H

Sualeh Asif - 3 years, 3 months ago

Log in to reply

Here is the general problem and solution by Xuming Liang:

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

Here are two solutions by him:

Serbian Problem

Sualeh Asif - 3 years, 3 months ago

Log in to reply

1 week,12 problems.

Prince Loomba - 3 years, 3 months ago

Log in to reply

Problem 13:

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

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

Sualeh Asif - 3 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...