Geometry for IOQM and RMO | An Excursion in Math Olympiad | Part 1

This is the first part of the complete story of Geometry for IOQM, RMO, AMC 10 and other similar Math Olympiads. Instead of going through a list of theorems and formulas we actually look at this as connected story meandering through different beautiful results and configurations.

#ioqm #matholympiadpreparation #amc10 #ukmt

[Krishiv_mohan]

5:45 I am Krishiv Mohan, a student at Cheenta and I would like to prove the theorem .
First, construct a point D such that AD and CD are perpendicular
For the proof, I will consider theta to be @.

AC^2=AB^2+BC^2-2.AB.BC.cos@
PROVED.
😊

[yadusgardenarulmurugan]

It would be useful for other exams also thank you sir 🙏🙏🙏🙏🙏🙏🙏

[geetikagrovermukhija]

Hello Sir! I liked the video a lot! Though the concepts were really easy, it was a nice revision of all the small residues that also had to be cleared up. I am Raghav Mukhija, a Level 6U student in Cheenta. Here is my solution to one of the questions asked:Δ

To Prove: The corresponding angles of an isosceles Δ are equal.

Consider the isosceles ΔABC, where AB = AC.
Construct a perpendicular from AD to side BC.

Consider ΔADB and ΔADC,
Given:
AB = AC (Isosceles Δ)
AD = AD (Common Arm)
∠ADB = ∠ADC = 90° (Perpendicular)
Hence, ΔADB ≅ ΔADC.

Thus, ∠ABD = ∠ACD (Corresponding Parts of Corresponding Triangles OR CPCT)

Hence, Proved!

[geetikagrovermukhija]

Another Proof:

Construction: Right ΔABC, with BX as the perpendicular at AC.

To Prove: ΔABX ~ ΔBCX ~ ΔABC

∠AXB = ∠ABC = ∠BXC = 90°
∠BAX = ∠BAC & ∠BCX = ∠BCA (Common Angles)

Hence, proving ΔABX ~ ΔBCX ~ ΔABC

[Bittú-hp]

I really want to take tution from cheenta but I can't afford the fees 😢