Geometry Marathon v1.0

[Moon]

I think that we can start a marathon here. The rule is very simple. I start this marathon with a problem. If someone solve it, s/he must submit another problem. This way the marathon will go on. The difficulty level of the problems should be around National Olympiad medium level.

Here goes the first problem:

Problem 1:
Let $ABC$ an equilateral triangle and $\Gamma$ its inscribed circle. If $D$ and E are points in the sides $AB$ and $AC$ respectively, such that $DE$ is tangent to $\Gamma$, show that
\[\frac{AD}{DB}+\frac{AE}{EC}=1\]

[Zzzz]

I have got a solution. I have used some identities.

Here are the identities:

i) The inradius of an equilateral triangle, $r=\frac{\sqrt{3}}{6} a$, here $a$ is the length of each side.
ii) If $ABC$ is a triangle and the excircle on side $BC$ touches $AB$ at $D$ then $AD=s$.
iii) If $ABC$ is a triangle, the radius of the excircle on side $BC,r_a=\frac{(ABC)}{s-a}$.

Now, let $AD=x, DE=z, AE=y$. $AB=BC=AC=a$. Radius of $\Gamma=R$. $\Gamma$ touches $AB$ at $Q$.

Using identity (ii), Semiperimeter of $ADE,s=AQ=\frac{1}{2}a$.
Using identity (i) and (iii), $R=\frac{\sqrt{3}}{6} a=\frac{(ADE)}{s-z}$
\[\Rightarrow (ADE)=\frac{\sqrt{3}a(a-2z)}{12}\]

Now,
\[\frac{AD}{BD}+\frac{AE}{EC}=1\]
\[\Leftrightarrow \frac{x}{a-x}+\frac{y}{a-y}=1\]
\[\Leftrightarrow 2a(x+y)-a^2=3xy\]
\[\Leftrightarrow 2a(2s-z)-a^2=3xy\]
\[\Leftrightarrow 2a(a-z)-a^2=3xy \]
\[\Leftrightarrow a^2-2az=4\sqrt{3}(ADE) [(ADE)=\frac{1}{2}xy\sin\angle DAE, \angle DAE=60^{\circ}]\]
\[\Leftrightarrow a^2-2az=4\sqrt{3}\cdot\frac{\sqrt{3}a(a-2z)}{12}\]
\[\Leftrightarrow a^2-2az=a^2-2az\]

Thanks for the problem

[Zzzz]

Ops.. I can't remember a cute problem Try this easy one and give a solution with a problem

Problem 2:
In a triangle $ABC, \angle ACB= 90^{\circ}. BC>AC.$ The perpendicular bisector of $AB$ intersects $BC$ at $D$ and $AC$ at $E$. $DE=AB$.

Find $\angle ABC$

[Mod Edit: Write problem number before each new marathon problem]

[TIUrmi]

Solution to problem 2:

Let \[\angle ABC = \Theta \]
As \[\angle DCE =\angle BME = 90^{\circ} \] $ E, C, M , B$ are concyclic where $M$ is the midpoint of $AB$.
Therefore, \[\angle CED = \angle MBD = \angle DAM = \Theta\]
and \[\angle CDE = \angle ADE = \angle BCA = 90^{\circ} - \Theta\]

From the fact that \[\triangle ABC \cong \triangle EDC\] follows that $CD = AC$ and so \[\angle ADC = \angle DAC = 90^{\circ} - 2\Theta\]

\[\angle CDE + \angle ADM + \angle ADC = 2(90^{\circ} - \Theta) + (90^{\circ} - 2\Theta) = 180^
{\circ}\]

\[\Theta = 22.5^{\circ}\]

[TIUrmi]

I have a question regarding problem 1.
Aren't $P$, $Q$, $R$ collinear? Here $P$ is the intersecting point of $CD$ and $BE$ and $Q$, $R$ are midpoints of $AB$ and $AC$ respectively? I couldn't prove this.


Problem 3:
In $\triangle ABC$, let the bisector of angle $A$ meet $BC$ at $U$. Prove that the perpendicular bisector of $AU$, the perpendicular to $BC$ at $U$, and the circumdiameter through $A$ are concurrent.

[Zzzz]

Let the circumcenter is $O$. $AO$ intersects the perpendicular line at $U$ at point $X$. Extended $AU$ meets circumcircle at $Y$. So, $Y$ is the midpoint of arc $BC$ and $OY\perp BC$.
From $\Delta OAY, OA=OY$. So, \[\angle OAY=\angle OYA\]
Again, $UX\parallel OY$. So, \[\angle OYA= \angle XUA\]
\[\Rightarrow \angle OAY= \angle XUA\]
From $\Delta XAU, XA=XU$
That means $X$ is equidistant from $A$ and $U$. Hence $X$ is a point on perpendicular bisector of $AU$.

qed

[Zzzz]

Problem# 4
$P$ is a point in the interior of $\Delta ABC$. The three lines through $P$ parallel to the sides of the triangle divide the triangle into three parallelograms and three triangles. Prove that, the perimeter of each of the three small triangles is equal to the length of the adjacent side iff $P$ is the incenter.

[Moon]

Cool! This topic is going to be the encyclopedia of nice problems!

Keep it up! I hope that more members will join. Where is Pranon and Promee???

[ishfaqhaque]

........................

[Zzzz]

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