Geometry Set

The problems are in no particular order. Please reshare the note if you like what you see, feel free to share comments or ideas and enjoy.

  1. Let PP and QQ be points on the side ABAB of the triangle ABCABC (with PP between AA and QQ) such that ACP=PCQ=QCB\angle{ACP}=\angle{PCQ}=\angle{QCB}, and let ADAD be the angle bisector of BAC\angle{BAC} with DD on BCBC. Line ADAD meets lines CPCP and CQCQ at MM and NN respectively. Given that PN=CDPN=CD and 3BAC=2BCA3\angle{BAC}=2\angle{BCA}, prove that triangles CQDCQD and QNBQNB have the same area. (Belarus 1999)

  2. Let ABCABC be a triangle and let ω\omega be its incircle. Let D1D_1 and E1E_1 be the tangency points of ω\omega with sides BCBC and ACAC, respectively. Let D2D_2 and E2E_2 be points on sides BCBC and ACAC, respectively, such that CD2=BD1CD_2=BD_1 and CE2=AE1CE_2=AE_1, and let PP be the intersection point of segments AD2AD_2 and BE2BE_2. Circle ω\omega intersects segment AD2AD_2 in two points, being the closest one to AA denoted by QQ. Prove that AQ=D2PAQ=D_2P.

  3. Points A,B,C,DA, B, C, D are four consecutive vertices of a regular polygon such that 1AB=1AC+1AD\dfrac{1}{AB}=\dfrac{1}{AC}+\dfrac{1}{AD}. How many sides does the polygon have?

Note by José Marín Guzmán
4 years, 10 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

3)The question number 3 is easy to solve with complex numbers.But also found a geometric solution.

Let the polygon be nn- sided.

Suppose the vertex after DD be EE.Hence the 55 consecutive vertices are A,B,C,D,EA,B,C,D,E ( arranged anticlockwise from A to E,say)

Firstly,1AB=1AC+1AD\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD} gives

AC.AD=AB.AD+AB.ACAC.AD=AB.AD+AB.AC...(1)(1)

One can easily show that quadrilateral ACDEACDE is cyclic.

So,by Ptolemy's theorem, AE.CD+AC.ED=AD.ECAE.CD+AC.ED=AD.EC...(2)(2)

Also,since the polygon is regular,we have AB=CD=EDAB=CD=ED and EC=ACEC=AC.

So,from (1)(1), AB.AD+AB.AC=AC.AD=CE.AD=AE.CD+AC.ED=AE.AB+AC.ABAB.AD+AB.AC=AC.AD=CE.AD=AE.CD+AC.ED=AE.AB+AC.AB

Hence,AD=AEAD=AE.

Since,these two diagonals are equal,number of vertices between AA and EE(moving clockwise) equals the number of vertices between AA and DD(moving anti-clockwise).

Hence,n5=2n-5=2 or n=7n=7.

Souryajit Roy - 4 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...