$\frac{HQ}{HF} = \frac{HE}{HA}$ $\frac{HD}{HR} = \frac{AD}{AP} = \frac{AB}{AE} = \frac{HC}{HE}$ $\frac{HF}{HD} = \frac{HA}{HC}$ Multiplying we get $:\ \ \frac{HQ}{HR} = 1$

By the cyclicity of $BEHD$ and $CFHD$, it follows that $\angle{EDH}=\angle{ABH}=90^\circ -\angle{A}=\angle{ACH}=\angle{FDH}$ and hence $H$ lies on the bisector of $\angle{EDF}$. By the same argument, $H$ lies on the bisectors of $\angle{DEF}$ and $\angle{DFE}$ which implies that $H$ is the incenter of triangle $DEF$. Further, since $AE \perp HE$ and $AF \perp HF$, it follows that $A$ lies on the external bisectors of $\angle{DEF}$ and $\angle{DFE}$ and hence is the $D$-excenter of triangle $DEF$. Therefore $P$ and $Q$ are the points of tangency of the incircle and excircle to side $EF$ in triangle $DEF$. Now let $DP$ intersect the incircle of $DEF$ at $R'$. Since the homothety with center $D$ that takes the incircle of $DEF$ to the $D$-excircle takes $R'$ to $P$, the tangent to the incircle of $DEF$ at $R'$ is parallel to $EF$. This implies that $R'$ is diametrically opposite to $Q$ and hence lies on $HQ$. Hence $R'=R$ and $HQ=HR$ since both lengths are the radius of the incircle of $DEF$.

The problem is a well known problem applied in a different cofiguration. Here is the well-known problem: Let $ABC$ be a triangle, $I_A$ the $A$-excircle. Let $P$ be the foot of perpendicular from $I_A$ to $BC$. Also, let $I$ be the incenter of $\triangle ABC$ and $S$ the foot of perpendicular from $ I$ to $BC$. Let $ T$ be symmetrical of $ S$ with respect to $I$. Then $A,T,P$ are collinear. The problem is just this lemma applied to $\triangle DEF$.

This problem was proposed by me. Yes, it can be solved by applying that famous lemma to the triangle $DEF$.

Actually a tricky solution with similarity would work quite well with this problem. Obviously $\triangle HEF \sim \triangle HCB$ , Q and D are corresponding points on similar figures, so $\frac{HQ}{HD}=\frac{EF}{BC}$, and $RQ//AP$ implies that $\frac{HR}{DH}=\frac{AP}{AD}$ Point P and D are corresponding points on similar figures $\triangle AEF$ and $\triangle ABC$ , so $\frac{AP}{AD}=\frac{EF}{BC}$ , hence $HQ=HR$. Lots of "Corresponding points on similar figures" are proposed in USA maths contests, who tells me why, I found them at least in USAMO 2009 and USA-TST 2011/2012 (>_<)

How many points the team has been entered this year?

My solution: By similar triangles , we have $\frac{HR}{AP}=\frac{DH}{AD}$ and $\frac{HQ}{AP}=\frac{HX}{XA}$. We thus show that $D$ and $X$ divide $AH$ in the same ratio externally and internally and it follows as a result that $HQ=HR$. Consider $\Delta AHB$ and line $FE$ and apply Menelaus' theorem to get $\frac{HX}{XA}\frac{AF}{FB}\frac{BE}{EH}=1$ and then a Ceva's theorem applied to the same triangle and Point $C$ yields : $\frac{HD}{AD}\frac{AF}{FB}\frac{BE}{EH}=1$, whence it follows what was sought. $\Box$

Proof is on similar lines to the ones previously posted. Let $H'$ be orthocentre of $AEF$. $AH'H\sim APD$ and so, $HH'||PD$. But easy to see that $HH'$ passes through midpoint of $PQ$ and hence, $H$ is midpoint of $QR$.

in complete $BFEC$ we have that $A,S,H,D$ are harmonic and now projecting through $P$ since $AP||HQ$ ie $HQ=HR$

It is kind of obvious from parallelism that $\dfrac{HQ}{HR}=\dfrac{HM}{AM} \cdot \dfrac{AD}{HD}$, where $\{ M \}=AD\cap EF$. In the same time, we have $\widehat{AEH}=90^\circ,\ \widehat{MEH}\equiv \widehat{HCD}\equiv \widehat{ HED}$ so $(A,H,M,D)$ is a harmonic division, i.e. $\dfrac{HQ}{HR}=1$.

It is well-known that $ H $ and $ A $ are the incenter and $ D $-excenter of $ \triangle{DEF} $ respectively. This implies that $ Q $ is the tangency point of the incircle of $ \triangle{DEF} $ with $ EF $ and that $ P $ is the tangency point of the $ D $-excircle of $ \triangle{DEF} $ with $ EF $. Letting $ Q' $ be the antipode of $ Q $ with respect to the incircle of $ \triangle{DEF} $ we have that $ D, Q', P $ are collinear (this is a standard Yufei configuration) so $ R = Q' $. Therefore $ HQ = HR $ so the desired ratio is equal to $ \boxed{1} $.

easy for USA TST! my solution =it is easy to see that $HQ$ and $AP$ are parallel. thus $\frac{QH}{AP}=\frac{HT}{TA} $ and $\frac{HR}{AP}=\frac{DH}{DA}$ where $T$ is intersection of $EF,AD$. where $DH=ABcosBcotC$ and $AD=ABsinB$ now sine law yields with easy angle chasing that $\frac{DH}{DA}=\frac{cosBcosC}{sinBsinC}$ now sine law triangle $HTE,TEA$ yield with angle chasing using cyclic quad $BCEF,AFHE$ that $\frac{HT}{TA}=\frac{cosBcosC}{sinBsinC}$ and thus $\frac{HT}{TA}=\frac{DH}{DA}$ and hence $HR=HQ$ we are done !

Here's a quick solution I don't think anyone has posted yet. Note $AP$ is parallel to $QR$. Let $T = EF \cap BC$ and $S=AD \cap EF$. Then we have $HR=AP \cdot \frac{DH}{DA}$ and $QH = AP \cdot \frac{HS}{AS}$. It's well known that $(T, F, S, E)$ is harmonic, so $(D, H, S, A)$ is harmonic after taking the pencil at $C$. So it follows that $\frac{DH}{HS}=\frac{AD}{AS}$ and the desired answer is $1$.

Let $H'$ be the orthocenter of $\triangle AEF$. Notice that $\triangle ABC \sim \triangle AEF$, so $H'H \parallel DP$. However, it is well known that $M$ lies on $HH'$, so $MH\parallel PR$. However, since $A$ is an excenter and $H$ is the incenter of $\triangle DEF$, then it is immediate that $QM=MP$ and that $HQ/HR=1$. $\Box$

Define $FE \cap BC = G, FE \cap AD = S.$ Let $AP \cap QR = A_{\infty}.$ $$(G, D; B, C) \stackrel{E}{=} (S, D; H, A) \stackrel{P}{=} (Q, R; H, A_{\infty}),$$so $H$ is midpoint of $QR,$ done.

It is well known that $H$ is the incenter of $\triangle{DEF}$, and that $A$, $B$ and $C$ are the $D$, $E$ and $F$ excenters of triangle $\triangle{DEF}$ respectively. It follows that P is the point of tangency of the $D$ excircle and $FE$. Let $\Gamma$ be the circle with radius $QH$ and center $H$. Lemma: The point of tangency of the $D$ excircle to $EF$, the point diametrically opposite $Q$, with respect to $\Gamma$, and D are collinear

Since the point $R'$ is on $QH$, it follows from the lemma that $R'=QH \cap PD$. Thus $R'=R$, implying that $HQ=HR$, so the answer is $1$. $\square$

https://artofproblemsolving.com/community/c473124h1565787_diameter_of_incircle_and_midpoints_of_altitudes

We can also bary bash. We claim that the answer is $1$. View $\triangle DEF$ as the reference triangle with the excenter/orthocenter duality. Then $H$ is the incenter of $\triangle DEF$ and $P$ and $Q$ are the incircle and $A$-excircle touchpoints on $BC$. We will show that $H$ is the midpoint of $\overline{QR}$. Now employ barycentrics with $D=(1,0,0),E=(0,1,0),F=(0,0,1)$ and let $EF=a,FD=b,DE=c$. Then $I=(a:b:c),P=(0:s-b:s-c),Q=(0:s-c:s-b)$. Now we will compute the coordinates of $R$. Since it lies on cevian $DP$, it can be parameterized by $R=(t:s-b:s-c)$ for some $t\in\mathbb{R}$. But since $H,R,Q$ are collinear, we must have $$\begin{vmatrix}t&s-b&s-c\\ a&b&c\\ 0&s-c&s-b\end{vmatrix}=0.$$Expanding the determinant, we have \begin{align*} t[b(s-b)-c(s-c)]+(s-b)[-a(s-b)]+(s-c)[a(s-c)]&=0\\t[b(s-b)-c(s-c)]&=a[(s-b)^2-(s-c)^2]\\ t\cdot \frac12 [b(a-b+c)-c(a+b-c)]&=a(s-b+s-c)(s-b-s+c)\\ t\cdot\frac12 [(b-c)(a-b-c)]&=-a^2(b-c)\\ t&=\frac{a^2}{s-a}\end{align*}so $R=(a^2:(s-a)(s-b):(s-a)(s-c)).$ We can normalize $R$ by dividing each component by $a^2+(s-a)(s-b+s-c)=a(a+(s-a))=as$. It suffices to check that $\frac{Q+R}2=H \iff Q=2H-R$. To do this, we will normalize each point, make a common denominator, and check each component. Normalization: \begin{align*} \bullet Q&=\left(0,\frac{s-c}{a},\frac{s-b}{a}\right)\\ \bullet H&=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)\\ \bullet R&=\left(\frac{a^2}{as}:\frac{(s-a)(s-b)}{as}:\frac{(s-a)(s-c)}{as}\right) \end{align*} Common denominator (we will pick $as$): \begin{align*} \bullet Q&=\left(0,\frac{s(s-c)}{as},\frac{s(s-b)}{as}\right)\\ \bullet I&=\left(\frac{a^2/2}{as},\frac{ab/2}{as},\frac{ac/2}{as}\right)\\ \bullet R&=\left(\frac{a^2}{as}:\frac{(s-a)(s-b)}{as}:\frac{(s-a)(s-c)}{as}\right) \end{align*} Checking $Q=2H-R$: \begin{align*} \bullet x&: 0=2(a^2/2)-a^2\\ \bullet y&: s(s-c)=2(ab/2)-(s-a)(s-b)\\ \bullet z&: s(s-b)=2(ac/2)-(s-a)(s-c) \end{align*} which is all true upon a little bit of expansion. Done.

The answer is $1$. Using basic properties of incenters, excenters, and the contact triangle, we may rewrite the problem as follows: ”New Problem” wrote: Let $DEF$ be a triangle with incenter $H$, $D$-incircle contact point $Q$, $D$-excircle contact point $P$. Prove that the $Q$-antipode on the incircle of $DEF$, $D$, and $P$ are collinear. This proves that the answer is $1$ because then $R$ would be said antipode and $HR=HQ=r$, where $r$ is the radius of the incircle of $\triangle DEF$. The proof of this is simple: Call the $Q$-antipode $R’$, draw $R’R’$ and let it meet $DE$ and $DF$ at $E’$ and $F’$ respectively. Then the homothety mapping $DE’F’$ to $DEF$ maps $R’$ to $P$, so $\overline{D-R’-P}$.

$(H,A; AD\cap EF, D)=-1, (Q,R; H,P_{\infty})=(AD\cap EF, D; H,A)=-1, \frac{HQ}{HR}=1$

We first use the following lemma. Lemma: Let $ABCD$ be a quadrilateral whose diagonals meet at $K$. Lines $AD$ and $BC$ meet at $L$ and line $KL$ meets $\overline{AB}$ and $\overline{CD}$ at $M$ and $N$ respectively. Then $(K, L; M, N)$ is a harmonic bundle. Proof: Let $P = \overline{AB} \cap \overline{CD}$ and let $Q = \overline{PK} \cap \overline{BC}$. Then we have obviously that $(Q, L; B, C) = -1$. (Use Ceva and Menelaus theorem). Then projecting from $P$ we have that $-1 = (Q, L; B, C) \stackrel{P}{=} (K, L; M, N)$. $\square$ By the above lemma, we have that $(H, A; \overline{AD} \cap \overline{EF}, D) = -1$. Projecting through $P$ we find $(P_{\infty}, H; Q, R) = -1$, where $P_{\infty} = AP \cap QR$. Hence $H$ is the midpoint of $\overline{QR}$, so that $HQ/HR = 1$. $\blacksquare$

The answer is $1$. To see this, focus just on triangle $DEF$. As $H$ is the incenter and $A$ is the $D$-excenter, the points $Q$ and $P$ are the respective contact points of the incircle and $D$-excircle. So $R$ is the antipode of $Q$ along the incircle.

We have a lot of arbitrary intersections with a ton of points on lines, so we use projective. Notice AP//QR; then $$-1=(C,B;D,EF\cap BC)=^E(A,H;D,EF\cap AD)=^P(P_{\infty},H;R,Q),$$so H is the midpoint of QR, whence the ratio is 1, as desired. $\blacksquare$

Let $X = AH \cap FE$. Then as $AP \parallel QR$, \[-1 = (A, H; X, D) \stackrel{P \to \overline{QR}}{\longrightarrow} (P_{\infty}, H; Q, R).\]Hence $H$ is the midpoint of $QR$ so that $HQ/HR = 1$. $\blacksquare$

Let $X_C= \overline{ED} \cap \overline{AB}$ and $L = \overline{AD} \cap \overline{EF}$. Now, note that since $AP \perp EF \perp HQ$, we have $AP \parallel QH$.Then, we know that \[(AB;FX_C) \overset{E}{=}(AH;LD) \overset{P}{=} (P_{\infty}H;QR)\]But this means that $H$ must be the midpoint of $QR$. Thus, \[\frac{HQ}{HR} = 1\]

Consider $\triangle DEF$. Note that $H$ is its incenter and $\triangle ABC$ is its excentral triangle. Thus $P$ is the touchpoint of the $D$-excircle and $Q$ is the touchpoint of the incircle to $\overline{EF}$. Now let $Q'$ be the antipode of $Q$ on the incircle. Note that there is a homothety centered at $D$ mapping the $D$-excircle to the incircle, so this homothety maps $P$ to $Q'$, meaning $Q'$ lies on $DP$. Also, $Q'$ lies on $QH$ for obvious reasons. This implies that $DP\cap QH = Q'$, so $R = Q'$. Finally, $HQ = HQ'$, so \[ \frac{HQ}{HR} = \frac{HQ}{HQ'} = \frac{1}{1} = 1.\]

We claim that $\dfrac{HQ}{HR} = 1$. Let $X = AD \cap EF$. First, we have $AP \parallel HQ$, so similar ratios give $$\dfrac{HQ}{AP} = \dfrac{HX}{AX}$$We also have $HR \parallel AP$, so similar ratios again yield $$\dfrac{AP}{HR} = \dfrac{AD}{HD}$$Multiplying the two ratios above, we obtain $$\dfrac{HQ}{HR} = \dfrac{HX \cdot AD}{AX \cdot HD} = \dfrac{AD}{AX} \cdot \dfrac{HX}{HD}$$However, we know that $BFEC$ and $DHEC$ are cyclic quadrilaterals, so $\angle FED = \angle FCB = \angle BED$, and it follows that $EH$ is the internal angle bisector of $\angle FED$. Since $AE \perp EH$, $AE$ is the external bisector of $\angle FED$, and by the angle bisector theorem, we have $$\dfrac{EX}{ED} = \dfrac{HX}{HD} = \dfrac{AD}{AX}$$which is equivalent to $\dfrac{AD}{AX} = \dfrac{HD}{HX}$, or $\dfrac{AD}{AX} \cdot \dfrac{HX}{HD} = 1$, as desired. $\blacksquare$

Shift our perspective to triangle $\triangle DEF$, with $H$ being the incenter and $\triangle ABC$ the ex-central triangle. Then since $A$ is the $D$-excenter, $P$ is the $D$-extouch point and $Q$ the $D$-intouch point. It is a well known lemma that $PD \cap QH$ lies on the incircle so $HQ = HR$ which gives us the answer of $1$.

easy one just focus on $\triangle AHC$ and then apply melaneus and since HQ parallel to AP you just apply thales and you are done

In terms of the orthic triangle, $Q$ is the touch point of the incircle with $EF$ and $R$ is diametrically opposite to $Q$ in this circle due to the famous diameter of incircle lemma so $\frac{HQ}{HR}$ is trivially $1$.