, 2019

Solution:
Let $PU$, $PV$ be tangents from $P$ to the incircle. We will invoke the dual of the Desargues Involution Theorem, which states the following:
Given a point $P$ in the plane and four lines $\ell_1, \ell_2, \ell_3, \ell_4$, consider the set of conics tangent to all four lines. Then we define a function on the pencil of lines through $P$ by mapping one tangent from $P$ to each conic to the other. This map is well defined and is a projective involution, and in particular maps $PA \rightarrow PD$, $PB \rightarrow PE$, $PC \rightarrow PF$, where $ABCDEF$ is the complete quadrilateral given by the pairwise intersections of $\ell_1, \ell_2, \ell_3, \ell_4$.
An overview of the projective background behind the (Dual) Desargues Involution Theorem can be found here: https://www.scribd.com/document/384321704/Desargues-Involution-Theorem, and a proof can be found at https://www2.washjeff.edu/users/mwoltermann/Dorrie/63.pdf.
Now, we apply this to the point $P$ and the lines $AB$, $AC$, $BC$, $YZ$, to get that the pairs
$$(PU, PV), (PY, PB), (PZ, PC)$$
are swapped by some involution. But we know that the involution on lines through $P$ which rotates by $90^\circ$ swaps the latter two pairs, thus it must also swap the first one and $\angle UPV = 90^\circ$. It follows by equal tangents that $IUPV$ is a square, thus $IP = r\sqrt{2}$ where $r$ is the inradius of $ABC$. Since $r = \frac{2K}{a+b+c} = \frac{21\sqrt{15}/2}{21} = \frac{\sqrt{15}}{2}$, we have $IP = \frac{\sqrt{30}}{2}$.

Solution:
Let $H$ be the orthocenter of $ABC$.

Lemma. $HI^2 = 2r^2 - 4R^2 \cos A \cos B \cos C$, where $r$ is the inradius and $R$ is the circumradius.

Proof. This follows from barycentric coordinates or the general result that for a point $X$ in the plane,
$$a XA^2 + b XB^2 + c XC^2 = (a+b+c) XI^2 + a AI^2 + b BI^2 + c CI^2$$
which itself is a fact about vectors that follows from barycentric coordinates. This can also be computed directly using trigonometry.

Let $E = BH \cap AC$, $F = CH \cap AB$, then note that $B, P, E, Y$ are concyclic on the circle of diameter $BY$, and $C, P, F, Z$ are concyclic on the circle of diameter $CZ$. Let $Q$ be the second intersection of these circles. Since $BCYZ$ is a tangential quadrilateral, the midpoints of $BY$ and $CZ$ are collinear with $I$ (this is known as Newton's theorem), which implies that $IP = IQ$ by symmetry. Note that as $BH \cdot HE = CH \cdot HF$, $H$ lies on the radical axis of the two circles, which is $PQ$. Thus, if $IP = IQ = x$, $BH \cdot HE$ is the power of $H$ with respect to the circle centered at $I$ with radius $x$, which implies $BH \cdot HE = x^2 - HI^2$.

As with the first solution, we claim that $x = r\sqrt{2}$, which by the lemma is equivalent to $BH \cdot HE = 4R^2 \cos A \cos B \cos C$. Then note that
$$BH \cdot HE = BH \cdot CH \cos A = (2R \cos B)(2R \cos C) \cos A,$$
so our claim holds and we finish as with the first solution.

Note. Under the assumption that the problem is well-posed (the answer does not depend on the choice of $Y, Z$, or $P$), then here is an alternative method to obtain $IP = r\sqrt{2}$ by making convenient choices. Let $U$ be the point where $YZ$ is tangent to the incircle, and choose $U$ so that $IU \parallel BC$ (and therefore $YZ \perp BC$). Note that $YZ \cap BC$ is a valid choice for $P$, so assume that $P$ is the foot from $U$ to $BC$. If $D$ is the point where $BC$ is tangent to the incircle, then $IUPD$ is a square so $IP = r\sqrt{2}$. (This disregards the condition that $Y$ and $Z$ are in the interior of segments $AC$ and $AB$, but there is no reason to expect that this condition is important.)