# Prove that $HK$ passes through a fixed point.

Point $$A$$ lies on the perpendicular bisector of $$BC$$. $$M$$ and $$N$$ are points respectively in line segments $$AB$$ and $$AC$$ such that $$MN$$ is tangent to the incircle of $$ABC$$ at point $$H$$. $$MP, NQ \perp BC$$ ($$P,Q \in BC$$). The intersection of $$MQ$$ and $$NP$$ is point $$K$$. Prove that $$HK$$ passes through a fixed point.

It's the midpoint of $$BC$$, obviously, but how to prove it is more important, which I failed to do. This problem is adapted from a recent competition.

• This is own problem from HSGS Class 10 Test! – Tran Quang Hung Jun 10 '19 at 17:41
• Yup! Can you solve it? – Lê Thành Đạt Jun 11 '19 at 1:05

Ignoring vertex $$A$$, this becomes a problem on the inscriptable $$\square BCNM$$. Let $$L$$ be the midpoint of $$\overline{BC}$$. Also, let $$M'$$ (instead of $$P$$), $$N'$$ (instead of $$Q$$), $$H'$$, $$K'$$ be the projections of $$M$$, $$N$$, $$H$$, $$K$$ onto $$\overline{BC}$$. Let the tangent segments from $$B$$ and $$C$$ to the incircle have length $$d$$; let the tangent segments from $$M$$ and $$N$$ have length $$m$$ and $$n$$. Finally, define $$m' := |MM'|$$, $$n':=|NN'|$$, $$m'':=|M'L|$$, $$n'':=|N'L|$$. Without loss of generality, $$m\leq n$$ so that $$m'\leq n'$$.

Certainly, if $$m=n$$, then $$\overleftrightarrow{HK}$$ meets $$\overline{BC}$$ at $$L$$. For $$m \neq n$$, we'll prove that $$L$$ is on $$\overleftrightarrow{HK}$$ by showing $$\triangle HH'L\sim \triangle KK'L$$ via $$|HH'||K'L|=|KK'||H'L| \tag{\star}$$

Parallelism and proportionality rules tell us that $$\frac{|M'H'|}{|M'N'|}=\frac{|MH|}{|MN|}=\frac{m}{m+n} \qquad |HH'|=m'+\frac{m}{m+n}(n'-m')=\frac{m'n+mn'}{m+n} \tag{1}$$ The Crossed Ladders Theorem tells us that $$\frac{1}{|KK'|}=\frac{1}{m'}+\frac{1}{n'} \quad\to\quad |KK'| = \frac{m'n'}{m'+n'} \tag{2}$$ (and, in fact, $$K$$ is the midpoint of the extension of $$\overline{KK'}$$ that meets $$\overline{MN}$$), whereupon some proportional thinking then yields $$|M'K'|:|K'N'|=m':n'$$, so that we have $$\frac{|M'K'|}{|M'N'|}=\frac{m'}{m'+n'} \tag{3}$$ Therefore, \begin{align} |H'L|&=|M'L|-|M'H'| = m'' - \,\frac{m}{m+n}(m''+n'') \;= \frac{m''n-mn''}{m+n} \\[6pt] |K'L|&=|M'L|-|M'K'| = m'' - \frac{m'}{m'+n'}(m''+n'')=\frac{m''n'-m'n''}{m'+n'} \end{align} \tag{4} Substituting in $$(\star)$$, and clearing denominators, we need only verify that $$(m'n+mn')(m''n'-m'n'') = m'n'(m''n-mn'') \tag{5}$$ That is, $$\frac{m}{n}\cdot\frac{m''}{n''} = \left(\frac{m'}{n'}\right)^2 \tag{6}$$

It seems like there's a geometric mean argument to be made, but I'm not seeing it. So, writing $$\theta$$ for the common angle at $$B$$ and $$C$$, we have $$\frac{m}{n}\cdot\frac{d-(m+d)\cos\theta}{d-(n+d)\cos\theta} = \left(\frac{m+d}{n+d}\right)^2 \quad\to\quad (d+m)(d+n)\cos\theta = d^2 - m n \tag{7}$$ This same relation results (for $$\theta \neq 0$$) from the observation that there's a right triangle with hypotenuse $$|MN|$$ and legs $$|m'-n'|$$ and $$m''+n''$$. $$(m+n)^2 = (m'-n')^2 + (m''+n'')^2 \qquad\to\qquad (7) \tag{8}$$ This equality establishes $$(\star)$$ and completes the proof. $$\square$$

I believe there's a cleaner way to link $$(6)$$ and $$(8)$$ (or to demonstrate $$(6)$$ some other way) without having to show equality through $$(7)$$. Again, I'm not seeing it. Perhaps I'll return to this question.

• Thanks for the answer. I would still want to way to prove $(6)$ without calculus. – Lê Thành Đạt Jul 22 '19 at 13:57
• @LêThànhĐạt: Um ... I didn't use any calculus. – Blue Jul 22 '19 at 14:08
• I mean advanced geometry. Whether where $\sin$ and $\cos$ are learned. – Lê Thành Đạt Jul 22 '19 at 14:33
• @LêThànhĐạt: Sine and cosine are used here in their most-fundamental roles: shorthand for particular ratios in a right triangle. It's possible to work through my argument without referencing them, but at the cost of some simplicity. – Blue Jul 22 '19 at 19:12

Here's a quicker way to relation $$(7)$$ from my previous answer.

Reiterating the setup from the previuos answer: Ignoring $$A$$, this is a problem on inscriptable $$\square BCNM$$. Let $$L$$ be the midpoint of $$\overline{BC}$$, and let $$M'$$ and $$N'$$ be the projections of $$M$$ and $$N$$ onto $$\overline{BC}$$. Let the tangential segments from $$B$$ and $$C$$ to the incircle have length $$d$$, let the tangential segments from $$M$$ and $$N$$ have lengths $$m$$ and $$n$$, and let define $$m':=|MM'|$$ and $$n':=|NN'|$$. We may assume $$m\leq n$$ so that $$m'\leq n'$$.

If $$m=n$$, then clearly $$\overleftrightarrow{HK}$$ contains $$L$$. For $$m\neq n$$, let the extensions of $$\overline{BC}$$ and $$\overline{MN}$$ meet at $$X$$. We'll prove collinearity of $$H$$, $$K$$, $$L$$ using Menelaus' Theorem, verifying this product:

$$\left[\frac{XL}{LM'}\right] \cdot \left[\frac{M'K}{KN}\right]\cdot\left[\frac{NH}{HX}\right] \stackrel{?}{=} -1 \tag{\star}$$ where "$$[\frac{PQ}{RS}]$$" indicates a signed ratio of lengths $$|PQ|$$ and $$|RS|$$: the sign is positive if $$\overrightarrow{PQ}$$ and $$\overrightarrow{RS}$$ point the same way, and negative if they point opposite ways. In the configuration shown, betweenness guarantees that the product of the signed ratios will be negative, as desired; therefore, we'll simplify things by proving absolute-value version of the relation: $$\frac{|XL|}{|LM'|} \cdot \frac{|M'K|}{|KN|}\cdot \frac{|NH|}{|HX|} \stackrel{?}{=} 1 \tag{\star\star}$$ Since $$\overline{XL}$$ and $$\overline{HX}$$ are tangent segments from $$X$$, they are congruent: $$|XL|=|HX| \tag{1}$$ Since $$\triangle MM'K\sim\triangle N'NK$$ and $$\triangle MM'B\sim \triangle NN'C$$ we have $$\frac{|M'K|}{|KN|}=\frac{m'}{n'}=\frac{m+d}{n+d} \tag{2}$$ Finally, we clearly have $$|NH|=n \qquad\text{and}\qquad |LM'| = d - (m+d)\cos\theta \tag{3}$$ Substituting $$(1)$$, $$(2)$$, $$(3)$$ into $$(\star\star)$$ gives

$$\frac{1}{d-(m+d)\cos\theta} \cdot\frac{m+d}{n+d} \cdot\frac{n}{1} \stackrel{?}{=} 1 \qquad\to\qquad d^2-mn \stackrel{?}{=} (m+d)(n+d)\cos\theta \tag{4}$$

This is relation $$(7)$$ from the previous answer. As shown there, the equality holds due to a Pythagorean relation that I won't repeat here (because I removed a few of the relevant lengths from the figure). Thus, we have $$(\star\star)$$, which implies that $$H$$, $$K$$, $$L$$ are indeed collinear. $$\square$$

Note. $$H$$ can actually be anywhere on the incircle (barring degeneracies). The reader is invited to adapt the above argument accordingly; that said, a coordinate-based proof, albeit computationally cumbersome, can be performed that is completely agnostic as to the position of $$H$$.

We can analytically prove that the straight line spanned by $$HK$$ passes through the midpoint of $$BC$$ as follows. Let the side $$BC$$ of the triangle lies along $$x$$-axis, and the origin $$O$$ is the midpoint of $$BC$$. Let ther vertex $$A$$ is a point $$(0,a)$$ and the vertices $$B$$ and $$C$$ are points $$(-b,0)$$ and $$(b,0)$$ respectively ($$a,b>0$$). Let $$I=(0,r)$$ be the center of the incircle and $$H=(x_h,y_h)=I+r(\cos\varphi,\sin\varphi)$$. Let $$R$$ (resp. $$S$$) be the tangent point of the incircle to the side $$AB$$ (resp. $$AC$$) of the triangle $$ABC$$. Since the point $$H$$ belongs to the upper arc of the incircle spanned by the points $$R$$ and $$S$$, $$\theta\le\varphi\le\pi-\vartheta$$, where $$\tan\theta=\tfrac ba$$. Let $$M=(m,h_m)$$, $$P=(0,h_m)$$, $$N=(n,h_n)$$, $$Q=(0,h_n)$$. Collinearity of points $$A$$, $$B$$, and $$M$$ implies $$h_m=a\left(\tfrac {m}{b}+1\right)$$. Similarly, collinearity of points $$A$$, $$C$$, and $$N$$ implies $$h_n=a\left(\tfrac {-n}{b}+1\right)$$. Let $$K=(x_k,y_k)$$. Similarity of triangles $$KMP$$ and $$KQN$$ implies $$\tfrac{x_k-m}{h_m}=\tfrac{n-x_k}{h_n}$$, so $$x_k=\tfrac {nh_m+mh_n}{h_m+h_n}.$$ Considering an angle $$NPQ$$, we obtain $$\tfrac{x_k-m}{y_k}=\tfrac{|PQ|}{|NQ|}=\tfrac{n-m}{h_n}$$. Thus $$y_k=\frac {h_n(x_k-m)}{n-m}=\frac{h_n}{n-m}\left(\frac {nh_m+mh_n}{h_m+h_n}-m\right)=\frac {h_mh_n}{h_m+h_n}.$$ It remains to show that the points $$O$$, $$K$$, and $$H$$ are collinear, that is $$\frac{b}{a}\left(\frac{n}{b-n}+\frac{m}{m+b}\right)= \frac{n}{h_n}+\frac{m}{h_m}=\frac{x_k}{y_k}=\frac{x_h}{y_h}=\tfrac{\cos\varphi}{1+\sin\varphi}.$$ Since $$MH\perp IH$$ we have

$$(r\cos\varphi-m)r\cos\varphi+\left(r(1+\sin\varphi)- a\left(\tfrac {m}{b}+1\right)\right)r\sin\varphi =0$$

$$(r\cos\varphi-m)\cos\varphi+\left(r(1+\sin\varphi)- a\left(\tfrac {m}{b}+1\right)\right)\sin\varphi =0$$

$$br-bm\cos\varphi+rb\sin\varphi-am\sin\varphi-ab\sin\varphi=0$$

If $$b\cos\varphi+ a\sin\varphi =0$$ then $$HI\perp AB$$, so $$H=R$$. Then $$N=A$$, so $$Q=O$$ and formally $$M$$ can be any point of $$AB$$. But in order to have the straight line spanned by $$HK$$ passes through $$O$$, we restrict ourselves to $$M=R$$ in this case. Then $$H=M$$ and $$K$$ belongs to $$MO$$.

Below we shall assume that $$a\sin\varphi+ b\cos\varphi \ne 0$$. Then

$$m=b\frac{r(1+\sin\varphi)-a\sin\varphi }{b\cos\varphi+ a\sin\varphi}.$$

It follows

$$\frac{m}{m+b}=1-\frac{ a\sin\varphi+ b\cos\varphi}{b\cos\varphi+ r(1+\sin\varphi)}.$$

Similarly, since $$NH\perp IH$$ we have

$$(r\cos\varphi-n)r\cos\varphi+\left(r(1+\sin\varphi)-a\left(\tfrac{-n}{b}+1\right)\right)r\sin\varphi =0$$

$$br-bn\cos\varphi+rb\sin\varphi+an\sin\varphi-ab\sin\varphi=0$$

If $$b\cos\varphi-a\sin\varphi=0$$ then $$HI\perp AC$$, so $$H=S$$. Then $$M=A$$, so $$P=O$$ and formally $$N$$ can be any point of $$AC$$. But in order to have the straight line spanned by $$HK$$ passes through $$O$$, we restrict ourselves to $$M=S$$ in this case. Then $$H=M$$ and $$K$$ belongs to $$NO$$.

Below we shall assume that $$b\cos\varphi-a\sin\varphi\ne 0$$. Then

$$n=b\frac{r(1+\sin\varphi) -a\sin\varphi}{b\cos\varphi-a\sin\varphi }.$$

It follows

$$\frac{n}{b-n}=-1+\frac{b\cos\varphi-a\sin\varphi }{b\cos\varphi-r(1+\sin\varphi)}.$$

Thus

$$\frac{m}{m+b}+\frac{n}{b-n}=\frac{b\cos\varphi-a\sin\varphi }{b\cos\varphi-r(1+\sin\varphi)}- \frac{ a\sin\varphi+ b\cos\varphi}{b\cos\varphi+ r(1+\sin\varphi)}=$$

$$\frac{2b\cos\varphi(r(1+\sin\varphi)-a\sin\varphi)}{b^2\cos^2\varphi-r^2(1+\sin\varphi)^2}$$.

So it remains to verify that

$$\frac ba\frac{2b\cos\varphi(r(1+\sin\varphi)-a\sin\varphi)}{b^2\cos^2\varphi-r^2(1+\sin\varphi)^2}=\frac{\cos\varphi}{1+\sin\varphi}$$

It suffices to check that

$$\frac{2b^2(r(1+\sin\varphi)-a\sin\varphi)}{b^2\cos^2\varphi-r^2(1+\sin\varphi)^2}=\frac{a}{1+\sin\varphi}$$

$$\frac{2b^2(r(1+\sin\varphi)-a\sin\varphi)}{b^2(1+\sin\varphi)(1-\sin\varphi)-r^2(1+\sin\varphi)^2}=\frac{a}{1+\sin\varphi}$$

$$2b^2(r(1+\sin\varphi)-a\sin\varphi)=a(b^2(1-\sin\varphi)-r^2(1+\sin\varphi))$$

$$(2b^2r-ab^2+ar^2)(1-\sin\varphi)=0$$

That is, we need to show that $$2b^2r-ab^2+ar^2=0$$. That is $$r=\tfrac{-2b^2\pm \sqrt{4b^4+4a^2b^2}}{2a}=\tfrac{-b^2\pm b\sqrt{b^2+a^2}}{a}$$. Indeed, the radius $$r$$ equals the area $$ab$$ of the triangle $$ABC$$ divided by its semiperimeter $$b+\sqrt{a^2+b^2}$$. It follows $$r=\frac ba\left(\sqrt{a^2+b^2}-b \right)$$.