Synthetic Geometry Group - Alan Yan's Proposal

Note that although these questions are marked as "Synthetic", you are allowed to solve them with any method you want.

  1. (Medium Difficulty) Let ABCABC be a triangle and let ω \omega be its incircle. Denote by D1D_1 and E1E_1 the points where ω\omega is tangent to sides BCBC and ACAC, respectively. Denote by D2D_2 and E2E_2 the points on sides BC and AC, respectively, such that CD2=BD1CD_2 = BD_1 and CE2=AE1CE_2 = AE_1, and denote by PP the point of intersection of segments AD2AD_2 and BE2BE_2. Circle ω\omega intersects segment AD2AD_2 at two points, the closer of which to the vertex A is denoted by Q. Prove that AQ=D2PAQ = D_2P.

  2. (Easy Difficulty) Let triangle ABCABC satisfy 2BC=AB+AC2BC = AB + AC and have incenter I and circumcircle ω\omega. Let D be the intersection of AIAI and ω\omega (with A, D distinct). Prove that I is the midpoint of AD.

  3. (Hard Difficulty) Let the incircle ω \omega of triangle ABCABC touch BC,CA,BC, CA, and ABAB at D,ED , E and FF, respectively. Let Y1,Y2,Z1,Z2Y_1 , Y_2, Z_1, Z_2 and MM be the midpoints of BF,BD,CE,CDBF , BD, CE, CD and BCBC, respectively. Let the intersection of Y1Y2Y_1Y_2 and Z1Z2Z_1Z_2 be XX. Prove that MX⊥BCMX \perp BC .

If you want solutions posted, just comment down below.

#Geometry #Brilliant #SyntheticGeometry #Pro

Note by Alan Yan
5 years, 8 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} ai−1 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 sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Q2. It is clear that ABDCABDC is cyclic. ∠DBC=∠DAC=∠DAB=∠DCB\angle DBC = \angle DAC = \angle DAB = \angle DCB. So, ΔDCB\Delta DCB is isosceles and DC=DBDC = DB. But, ∠DIB=∠DBI\angle DIB = \angle DBI(by some angle chasing). So, DI=DB=DCDI = DB =DC. But, since ABDCABDC is cyclic, using ptolemy's theorem we get AD=2DB=2DIAD = 2DB = 2 DI, which implies that AI=IDAI = ID. So, II is mid-point of ADAD.

Surya Prakash - 5 years, 8 months ago

Does anyone want solutions?

Alan Yan - 5 years, 8 months ago

Log in to reply

yes i want soutions please post the solutions.

onkar tiwari - 5 years, 5 months ago
×

Problem Loading...

Note Loading...

Set Loading...