HMMT-Feb 2019 guts, Problem 31

Let $A B C$ be a triangle with $A B=6, A C=7, B C=8$. Let $I$ be the incenter of $A B C$. Points $Z$ and $Y$ lie on the interior of segments $A B$ and $A C$ respectively such that $Y Z$ is tangent to the incircle. Given point $P$ such that \[ \angle Z P C=\angle Y P B=90^{\circ} \] find the length of $I P$.

Solution: Solution 1. Let $P U, P V$ tangent 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 $P A \rightarrow P D, P B \rightarrow P E, P C \rightarrow P F$, where $A B C D E F$ 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: \href{https://www.scribd.com/document/384321704/Desargues-Involution-Theorem}{https://www.scribd.com/document/384321704/Desargues-Involution-Theorem}, and a proof can be found at \href{https://www2.washjeff.edu/users/mwoltermann/Dorrie/63.pdf}{https://www2.washjeff.edu/users/mwoltermann/Dorrie/63.pdf}. Now, we apply this to the point $P$ and the lines $A B, A C, B C, Y Z$, to get that the pairs \[ (P U, P V),(P Y, P B),(P Z, P C) \] 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 $\angle U P V=90$. It follows by equal tangents that $I U P V$ is a square, thus $I P=r \sqrt{2}$ where $r$ is the inradius of $A B C$. Since $r=\frac{2 K}{a+b+c}=\frac{21 \sqrt{15} / 2}{21}=\frac{\sqrt{15}}{2}$, we have $I P=\frac{\sqrt{30}}{2}$. Solution 2. Let $H$ be the orthocenter of $A B C$. Lemma. $H I^{2}=2 r^{2}-4 R^{2} \cos (A) \cos (B) \cos (C)$, 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, \[ a X A^{2}+b X B^{2}+c X C^{2}=(a+b+c) X I^{2}+a A I^{2}+b B I^{2}+c C I^{2} \] which itself is a fact about vectors that follows from barycentric coordinates. This can also be computed directly using trigonometry. Let $E=B H \cap A C, F=C H \cap A B$, then note that $B, P, E, Y$ are concyclic on the circle of diameter $B Y$, and $C, P, F, Z$ are concyclic on the circle of diameter $C Z$. Let $Q$ be the second intersection of these circles. Since $B C Y Z$ is a tangential quadrilateral, the midpoints of $B Y$ and $C Z$ are collinear with $I$ (this is known as Newton's theorem), which implies that $I P=I Q$ by symmetry. Note that as $B H \cdot H E=C H \cdot H F, H$ lies on the radical axis of the two circles, which is $P Q$. Thus, if $I P=I Q=x$, $B H \cdot H E$ is the power of $H$ with respect to the circle centered at $I$ with radius $x$, which implies $B H \cdot H E=x^{2}-H I^{2}$. As with the first solution, we claim that $x=r \sqrt{2}$, which by the lemma is equivalent to $B H \cdot H E=$ $4 R^{2} \cos (A) \cos (B) \cos (C)$. Then note that \[ B H \cdot H E=B H \cdot C H \cos (A)=(2 R \cos (B))(2 R \cos (C)) \cos (A) \] 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 $I P=r \sqrt{2}$ by making convenient choices. Let $U$ be the point where $Y Z$ is tangent to the incircle, and choose $U$ so that $I U \| B C$ (and therefore $Y Z \perp B C)$. Note that $Y Z \cap B C$ is a valid choice for $P$, so assume that $P$ is the foot from $U$ to $B C$. If $D$ is the point where $B C$ is tangent to the incircle, then $I U P D$ is a square so $I P=r \sqrt{2}$. (This disregards the condition that $Y$ and $Z$ are in the interior of segments $A C$ and $A B$, but there is no reason to expect that this condition is important.) $\fbox{\frac{\sqrt{30}}{2}}$.

Answer: \frac{\sqrt{30}}{2}