User talk:Mathboy100

Revision as of 18:18, 28 January 2024 by Ddk001 (talk | contribs) (SO RESPECT THE RULES.)

THIS IS MY TALK.

Not yours.

SO RESPECT THE RULES.

jkjkjk just like chat here and share problems i guess

Good we can share problems!

$\Delta ABC$ is an isosceles triangle where $CB=CA$. Let the circumcircle of $\Delta ABC$ be $\Omega$. Then, there is a point $E$ and a point $D$ on circle $\Omega$ such that $AD$ and $AB$ trisects $\angle CAE$ and $BE<AE$, and point $D$ lies on minor arc $BC$. Point $F$ is chosen on segment $AD$ such that $CF$ is one of the altitudes of $\Delta ACD$. Ray $CF$ intersects $\Omega$ at point $G$ (not $C$) and is extended past $G$ to point $I$, and $IG=AC$. Point $H$ is also on $\Omega$ and $AH=GI<HB$. Let the perpendicular bisector of $BC$ and $AC$ intersect at $O$. Let $J$ be a point such that $OJ$ is both equal to $OA$ (in length) and is perpendicular to $IJ$ and $J$ is on the same side of $CI$ as $A$. Let $O’$ be the reflection of point $O$ over line $IJ$. There exist a circle $\Omega_1$ centered at $I$ and tangent to $\Omega$ at point $K$. $IO’$ intersect $\Omega_1$ at $L$. Now suppose $O’G$ intersects $\Omega$ at one distinct point, and $O’, G$, and $K$ are collinear. If $IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2$, then $\frac{EH}{BH}$ can be expressed in the form $\frac{\sqrt{b}}{a} (\sqrt{c} + d)$, where $b$ and $c$ are not divisible by the squares of any prime. Find $a^2+b^2+c^2+d^2+abcd$.

Someone mind making a diagram for this?

~Ddk001