Problem 3! IMO 2015

Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma$ be its cirumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$ . Let $M$ be the midpoint of $BC$ . Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$ . Assume that the points $A$ , $B$ , $C$ , $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

This is part of the set IMO 2015

#Geometry #Circumcenter #Orthocenter #TangentOfCircles #Circumcircles

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.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered 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"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I can imagine that inversion kills this problem, here's my solution without using that, I don't expect many to understand this at this point as I wrote it in a hurry.

Let $I$ denote the midpoint of $HQ$ . It suffices to show that the circumcenter of $KFM$ lies on $IK$ , which equates to proving $\angle IKM=\angle 90-\angle KFC=\angle KFH$ .

$AH,QH,KH$ are extended to meet $(O)$ again at $E,Z,X$ respectively. It is well known and obvious that $O\in QX, AZ, M\in QZ$ . Let $F'$ denote the midpoint of $XH$ , $AF'\cap (O)=Y$ .

Since $(XZ\parallel MF')\perp (HQ$ , therefore $XZ\parallel MF'\parallel AQ$ . By the converse of Pascal's theorem, we can deduce that $AF'\cap KM=Y$ . Note that $\triangle AHXF'\sim \triangle KHEF$ , thus $\angle FKE=\angle XAF'=\angle XKY$ . Additionally, $\angle IKH=\angle OKA=\angle AZK=\angle AEK$ by spiral similarity about $K$ between $(I),(O)$ . Hence $\angle HFK=\angle AEK+\angle FKE=\angle IKH+\angle XKY=\angle IKM$ and we are done.

Xuming Liang - 5 years, 11 months ago

I don't know if that works or not, but try to use Nine-point circle to prove it. And I think that the point $K$ is probably Feuerbach point of some triangle.

Samuraiwarm Tsunayoshi - 5 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$