Chilling Geometry

Geometry Level 5

Let $O$ be the circumcentre of an acute triangle $\Delta ABC$ , with sides equal to $5$ , $6$ , and $7$ units in length, and with orthocentre $H$ . Let $P$ be any point on the major arc $BC$ (the arc not containing point $A$ ) of the circumcircle of $\Delta ABC$ , except $B$ and $C$ . Let $D$ be a point outside $\Delta ABC$ such that $\overline{AD}=\overline{PC}$ and $AD || PC$ . Let $K$ be the orthocentre of $\Delta ACD$ . The distance of the circumcentre of $\Delta ABC$ to $K$ can be expressed as $\dfrac{a\sqrt{b}}{4c}$ , where $a,b,c \in \mathbb{Z^+}$ , $\gcd(a,c)=1$ and $b$ does not contain the square of any prime number in its prime factorization. Determine the value of $a+b+c$ .

The answer is 47.

1 solution

The Linh Nguyen
May 5, 2015

(Warning : the figure above does not fit in size)

We got that ADCP is a parallelogram so $CP // AD$ . K is the orthocenter of ACD triangle so $CK \perp AD$ then $CK \perp CP$ .

Similarly, we got $AK \perp AP$ .

Therefore, ABCK is a cyclic quadrilateral and $OK = OA = OB = OC = R$ (assume that R is the radius of the circumcircle of ABC triangle).

We can calculate the area of ABC triangle by using Heron's Formula : $S = 6\sqrt{6}$

But the area of ABC triangle can also be expressed as $S = \frac{AB \times BC \times AC}{4R}$ thus $R = \frac{35\sqrt{6}}{24}$ .

Hence, $a=35$ , $b=6$ , $c=6$ then $a + b + c = \boxed{47}$ .

Chilling Geometry

The answer is 47.

1 solution

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