Exam (elaborations)
IranIncenter
- Course
- Institution
Exam of 6 pages for the course Nursing 220 Final Exam Review at Nursing 220 Final Exam Review (IranIncenter)
[Show more]
Seller
Follow
modockochieng06
Content preview
Two Important Lemmas in Olympiad GeometryJeffrey Kwan
October 9, 2017
Introduction
In this article, I present the key lemmas for the legendary Iran TST 2009 problem 9,
which is famous enough to have an entire configuration named for it. I include the
midpoint of altitudes lemma and the right angle on incircle chord lemma, which are both
crucial to the Iran problem. There are also five practice problems at the end.
§1 Midpoint of Altitudes
The midpoint of altitudes configuration involves two key collinearities, described in the
following lemma.
Lemma 1.1 (Midpoint of Altitudes)
In triangle ABC with incenter I and A-excenter IA , let D and T be the incircle and
A-excircle tangency point with BC. If M is the midpoint of the A-altitude, then
M , I, T collinear and M , D, IA collinear.
Proof. The proof is not difficult, but it is slightly tricky. Here I will only prove M , I, T
collinear since the other one is analogous.
A
D′
M
I
D T
B C
A1
IA
1
, Jeffrey Kwan (October 9, 2017) 2 Incenter Perpendicularity Lemma
This first thing to note is that A, D0 , T collinear, where D0 is the antipode of D in the
incircle. Indeed, the homothety at A sending the incircle to the excircles maps D0 to T
since they are both “top points.”
Now consider the homothety at T sending A1 A to DD0 . Where does M go? Since it’s
the midpoint of AA1 , it is sent to the midpoint of DD0 , which is precisely I!
§2 Incenter Perpendicularity Lemma
This incenter perpendicularity is slightly trickier than the previous result.
Lemma 2.1 (Right Angle on Incircle Chord)
The incircle of 4ABC is tangent to BC, CA, AB at D, E, F respectively. Let M
and N be the midpoints of BC and AC. If K is the intersection of lines BI and
EF , then BK ⊥ KC. In addition, K lies on line M N .
Proof. The first part can be reduced to showing that pentagon CDIEK is cyclic. Luckily,
this is pretty simple. Since D and F are reflections across line BI, we have
∠KDC = 180◦ − ∠BDK = 180◦ − ∠BF K = ∠AF E = ∠AEF = ∠KEC,
which completes the proof.
A
E K
N
F
I
B D M C
For the second part, observe that M is the circumcenter of 4BKC. Thus ∠CM K =
2∠KBC = ∠B, so M K k AB. Since M N k AB, it follows that K lies on line M N .
§3 The Legendary Iran TST
Now we are finally ready to tackle the Iran TST problem. While the problem is quite
difficult without knowing such lemmas, the solution with them is actually quite short!
2
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller modockochieng06. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $7.99. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
77333 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now