perplexus.info :: Geometry : Right Triangles

Right Triangles (Posted on 2010-01-22)

Triangle ABC is right-angled at C, a = |BC|, b = |CA|, with a < b.
Squares ABDE, BCFG, and CAHI are erected externally to triangle ABC.
Lines HE and DG intersect at P, lines DG and FI intersect at Q, and lines FI and HE intersect at R.

If triangle PQR is right-angled, determine the value of b/a.

Submitted by Bractals

Rating: 5.0000 (1 votes)

Solution: (Hide)

Apply an x-y coordinate frame such that
A (b,0), B (0,a), C (0,0), D (a,a+b), E (a+b,b), F (-a,0), G (-a,a), H (b,-b), and I (0,-b)
Two lines are perpendicular if the product of their slopes is -1
slope(PQ) = slope(DG) = b/2a slope(QR) = slope(FI) = -b/a slope(RP) = slope(HE) = 2b/a slope(QR)*slope(RP) = 2b²/a² = -1 ==> a = b√2 ==><== a < b. slope(RP)*slope(PQ) = b²/a² ≠ -1. slope(PQ)*slope(QR) = -b²/2a² = -1 ==> b = a√2.
Therefore, b/a = √2.

See Jer's post for alternate methods using vectors and trigonometry.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Answer	K Sengupta	2024-03-19 09:28:17
re(2): Thank you Descartes. 3 solutions.	Jer	2010-01-24 02:24:13
re: Thank you Descartes. 3 solutions.	Bractals	2010-01-22 18:07:31
Thank you Descartes. 3 solutions.	Jer	2010-01-22 15:05:10
Using Geometer's Sketchpad ... (spoiler)	Charlie	2010-01-22 14:12:20

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