In this write-up I am going to find and prove the following:

Construct any acute triangle ABC and its circumcircle. Construct the three altitudes AD, BE, and CF. Extend each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.

Find AP/AD + BQ/BE + CR/CF.

GSP exploration shows that this sum is equal to 4.

To be able to prove it, I will first need to prove that

HD/AD + HE/BE + HF/CF = 1

Then, as a corollary I will prove that

AH/AD + BH/BE + CH/CF = 2

First of all, I will show that HD = DP, HE = EQ and HF = FR.

Observe that angles BPA and BCA are congruent because they are both inscribed angles seeing the same arc.

Then observe that angle BHD is also equal in measure to BPA and BCA. This is true because the two legs of BHD are respectively perpendicular to the two legs of BCA (BH perpendicular to BC and HD perpendicular to CA). This shows that BPH is an isosceles triangle and that BD, being an altitude, is also the median. Hence HD = DP. With a similar argument, HE = EQ and HF = FR.

Now going back to our problem, we can write

AP/AD + BQ/BE + CR/CF = (AD + DP)/AD + (BE + EQ)/BE + (CF + FR)/CF = AD/AD + DP/AD + BE/BE + EQ/BE + CF/CF + CR/CF = 3 + DP/AD + EQ/BE + FR/CF

Replace DP, EQ and FR with HD, HE and HF respectively:

3 + DP/AD + EQ/BE + FR/CF = 3 + HD/AD + HE/BE + HF/CF

Now, observe that the area of the triangle ABC is equal to

(AD.BC)/2 = (BE.AC)/2 = (CF.AB)/2 which is also equal to sum of the areas of the triangles

AHB, BHC and CHA:

(HD.BC)/2 + (HE.AC)/2 + (HF.AB)/2 = Area(ABC)

In the second equation replace BC with 2Area(ABC)/AD, AC with 2Area(ABC)/BE and AB with 2Area(ABC)/CF:

Area(ABC)[HD/AD + HE/BE + HF/CF] = Area(ABC)

and HD/AD + HE/BE + HF/CF = 1

Putting this back in

AP/AD + BQ/BE + CR/CF = 3 + HD/AD + HE/BE + HF/CF

gives

AP/AD + BQ/BE + CR/CF = 4

As a corollary observe that

HD/AD + HE/BE + HF/CF = (AD ­ AH)/AD + (BE ­ BH)/BE + (CF ­ CH)/CF = 3 - (AH/AD + BH/BE + CH/CF)

Hence, AH/AD + BH/BE + CH/CF = 2