The problem: Let A be an outside point from the circle with the center O. Draw the secant ABC of the mentioned circle. The tangents at B and C got an intersection, named K. From K, we draw the perpendicular to AO. Name the intersection H. E and F are 2 intersections of KH and the circle with the center O. (E is between K and F). We name M for the intersection of KO and BC. Prove that:
- There exists a circle that pass throught 4 points E, M, O and F
- AE and AF are tangents of the circle with center O
Notes:
- I tried to translate this problem from my language to English. So If I use wrong grammar structure or wrong expression, please forgive me.
- This is not my homework. I like to look for and solve "hard" problems, by my view. This is one of those, and I can't find out the way to solve it.
Note by
Đức Việt Lê
5 years, 4 months ago
No vote yet
5 votes
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestThis problem can be solved by the similarity relationships in a right triangle with the altitude of the hypotenuse. Here's a sketch of my solution: 1. By the power of a point theorem(or whatever you want to call it, it's basically simiarity), we get that KExKF=KB^2. Moreover, since ∠KBO=90° and BM⊥KO, we get KMxKO=KB^2=KExKF→O,M,E,F are concyclic. 2. We know that AE is tangent to circle O iff ∠OEA=90°. Since ∠KMA=∠KHA=90°, so K,M,H,A are concyclic. Thus OHxOA=OMxMK=OB^2=OE^2→By SAS similarity △EOH is similar to △AOE→∠OEA=∠EHO=90° and that proves AE is tangent to circle O. By analogy AF is also tangent to circle O.
Log in to reply
sorry for not following the formatting guidline :(
Log in to reply
Thanks a lot! I can understand the solution that you gave me. But I cannot figure it out :D
Log in to reply