JHMT 2018 Geometry Problem 10

2018 JHMT Geometry Problem 10

In an acute triangle $ABC$, the altitude from $C$ intersects $\overline{AB}$ at $E$ and the altitude from $B$ intersects $\overline{AC}$ at $D$. $\overline{CE}$ and $\overline{BD}$ intersect at a point $H$. A circle with diameter $\overline{DE}$ intersects $\overline{AB}$ and $\overline{AC}$ at points $F$ and $G$ respectively. $\overline{FG}$ and $\overline{AH}$ intersect at $K$. If $\overline{BC} = 25$, $\overline{BD} = 20$, and $\overline{BE} = 7$, the length of $\overline{AK}$ is of the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

Nine-point circle?

I’ll be honest. As I was reading this problem and I saw that there was a circle passing through feet of altitudes, I immediately thought it was going to be about the nine-point-circle. I also immediately thought that I was going to have no chance to solve this because I don’t know anything about the nine-point circle. I literally don’t even know what all nine points are - I only know six. Fortunately, this circle only goes through two of those.

So, not a nine-point circle. Whew!

Diagram

Let’s come up with a diagram:

Pythagorean Triples

The first thing I noticed was that there’s a $7$, a $20$, and a $25$, so definitely some Pythagorean triples. Indeed there’s a 7-24-25 and a 15-20-25 (a.k.a a 3-4-5 triangle in disguise). Let’s update our diagram, ignoring how not-to-scale it is:

Parts of the Altitudes

Next I tried to see if I could “solve” the triangle and find its side lengths and cevians (altitudes in this case). Similar triangles abound! Notably, $\triangle CDH \sim \triangle BEH$. Let $\overline{DH} = x$ and $\overline{CH} = y$, and so $\overline{EH} = 24 - y$. By similar triangles:

$$\begin{align*} \frac{\overline{DH}}{\overline{DC}} &= \frac{\overline{EH}}{\overline{BE}} \\[10pt] \text{or} \quad \frac{x}{15} &= \frac{24 - y}{7} \end{align*}$$

and then by Pythagorean theorem:

$$15^2 + x^2 = y^2.$$

Solving this quadratic system gives us:

$$x = \frac{45}{4}, \quad y = \frac{75}{4}$$

Solving the Side Lengths

Now that we know how the altitudes divide each other, I can use mass-points to find the remaining side lengths. Here’s the updated diagram showing just the triangle and its altitudes:

Mass-pointing:

Whoa, so $\overline{AD}=\overline{DC}$ and this triangle was isosceles the whole time? I wish I knew that before. I think this discovery is important enough to warrant a diagram re-draw.

Diagram Redo

You’ll notice I added some auxiliary lines - these will be important shortly.

Inscribed-angles

The really crazy part about this problem is that we’ve just created a mini-version of our original problem within our triangle! What does this mean? Note that $\angle EFD$ and $\angle EGD$ both inscribe a semi-circle, so they are actually right angles. Now, since basically triangle in this whole problem is similar to some other triangle, $\triangle ABC \sim \triangle ADE$, and the lines $EG$ and $FD$ correspond to the altitudes $CE$ and $BD$.

Let me try to highlight it so it’s a little easier to see:

This seems like a bizarre coincidence to me, but of course somebody stronger in geometry will probably just say it’s obvious. Oh well.

Final Fact

The final piece of information I found was that $\triangle AKF \sim \triangle AEH$. Obviously they share $\angle A$, but it was less obvious that $\angle AFK$ was the same as $\angle AHE$. First note that $\triangle AEH \sim \triangle EGD$. The next critical non-obvious fact is that both $\angle GFD$ and $\angle GED$ subtend the same arc $\overset{\frown}{GD}$ on our circle and are thus equal. Without going through all of the details, we can use this to show that $\angle AFK = \angle AHE$ and then that $\triangle AKF \sim \triangle AEH$ by AA similarity:

Putting It All Together

Whew! All of this, just to figure out these two triangles were similar? Is there a sexier way? I don’t know, I couldn’t see one. Either way, we’re essentially done. Remember $AF$ bisects our isosceles triangle, so $\overline{AF}=\overline{FE}=9$. We can also use the Pythagorean theorem to find the hypotenuse of our larger triangle,

$$AH = \sqrt{EH^2 + EA^2} = \sqrt{5.25^2 + 18^2} = \frac{75}{4}$$

Lastly, by similar triangles:

$$\begin{align*} \frac{AK}{AF} &= \frac{AE}{AH} \\ \Longrightarrow \quad AK &= AF \cdot \frac{AE}{AH} = 9\cdot\frac{18}{\frac{75}{4}} = \frac{216}{25} \end{align*}$$

Since the problem wants us to find the length of $\overline{AK}$ in the form $\frac{p}{q}$ and give $p + q$, the final answer is thus:

$$p + q = 216 + 25 = \boxed{241}$$

Final Thoughts

OK, so I massively underestimated this problem going into it. So much so that I’m nearly certain I’m missing some key idea(s) that will dramatically simplify this problem. Do you really need to solve for every segment of the original triangle and then use mass-points? It feels like there must be a simpler way to immediately realize our big triangle was actually isosceles without computing side lengths. But I don’t actually see it.

I also have a sneaking suspicion that there’s a cooler way to show that $\triangle AKF \sim \triangle AEH$. I mean, it’s sort of crazy to me that $\angle AKF$ is a right angle. Again, I feel like I’m missing some insight that would allow me to immediately arrive at that conclusion without even considering angle-chasing and using the inscribed angle theorem on arc $\overset{\frown}{GD}$.

Either way, this problem was a monster for me. I found out that the triangle was isosceles relatively quickly by finding all of the side lengths and altitude ratios as I showed above. However, from there I had a lot of trouble figuring out what was special about points $F$ and $G$ and what the intersection of $FG$ and $AH$ could possibly mean. I got so stuck I almost just abandoned my train of thought to switch to coordinate bashing, when I had the epiphany that there were probably only two types of similar triangles in this problem: triangles that look like $\triangle BEH$ and triangles that look like $\triangle AEH$. I assumed this to be the case and got an answer, but it was just a guess. I really couldn’t see why $\angle AKF$ should be a right angle. Actually showing this was true took me well over an hour :(

Better Solution (Update)

I asked some stronger mathematicians for help (thanks Justin and Tony), and they informed me I did indeed miss an idea. I’m naming it the FAST Lemma:

FAST Lemma

FAST for Feet of Altitude Similar Triangle.

In other words, if we connect two of the feet of the altitudes of any triangle, the resulting smaller triangle will be similar to the original. You can easily prove this by noting that, for example, quadrilateral $BCEF$ is cyclic and looking at the arcs subtended by $\angle C$ and $\angle B$‘s supplement.

Invoking FAST Lemma

We actually invoke the FAST Lemma twice in this problem. Once to show that $\triangle ABC\sim\triangle ADE$ and once again on $\triangle ADE$ to show that $\triangle AED\sim\triangle AFG$ since $EG$ and $FD$ are also altitudes of $\triangle ADE$!

I’ve tried to highlight everything below to make it easier to see:

Since all three triangles are similar, $FG\parallel BC$ and the following two triangles are similar:

Solving the Triangle

Since our cevians were actually altitudes, mass-points was overkill. Better to simply use trig to find $\sin A$:

$$\begin{align*} \sin A &= \sin(180^\circ - (B + C)) \\ &= \sin(B + C) \\ &= \sin B \cos C + \cos B \sin C \\ &= \left(\frac{24}{25} \cdot \frac{15}{25}\right) + \left(\frac{7}{25} \cdot \frac{20}{25}\right) \\ &= \frac{4}{5} \end{align*}$$

But wait, $\sin C=\frac{4}{5}$ as well, so our triangle is isosceles! We figured it out this time with barely any work too. And since $\triangle ABC\sim\triangle ADE$, $DF$ is a perpendicular bisector of $AE$ just like $BD$ is a perpendicular bisector of $AC$.

Using this information along with the same Pythagorean triple observations as above we instantly get the following side lengths:

By similar triangles:

$$AK = AF\cdot\frac{AJ}{AB} = 9\cdot\frac{24}{25} = \frac{216}{25}$$

which is the same answer as before.