## Thursday, July 9, 2009

### Problem 317: Right triangle and Inscribed Squares

Proposed Problem
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Complete Problem 317
Level: High School, SAT Prep, College geometry

1. Denoting AH & PC by c & d resply. and noting the fact that all rt. angled triangles within Tr. ABC are similar to it, we can say that c/(c+a)= a/x while d/(d+b)= b/x. Further, c/a = AB/BC and d/b = BC/AB or c/(c+a)= AB/(AB+BC) and d/(d+b)= BC/(AB+BC). This gives us, a/x + b/x = c/(c+a) + d/(d+b)= AB/(AB+BC) + BC/(AB+BC) = 1. In other words, x = a + b QED.

2. tr JKE and tr EBF are similar ( an JEK = an BFE )

=> EB/BF = a/x-a

tr EBF and MFN are similar ( an BFE = an FNM )

=> EB/BF = x-b/b

from above => a/x-a = x-b/b

=> x = a + b

3. May I ask a question?
ABC is a right angled Triangle with right angle at A. How to inscribe a square PQRS such that P. Q lie on the sides AB, AC respectively and both R, S lie on the hypotenuse BC ?.
Rajahmundry, INDIA.

1. Geometry solution:
https://photos.app.goo.gl/Hx32KiWgwVlZtzIM2
Let BC= a and AH= h ( see sketch)
1. Draw altitude AH= h
2. Extend BT such that BT= AH= h
3. Connect AT
4. Draw BQ//AT . Q is on AC
5. Draw QP//BC. P is on AB
6. Draw QR ⊥BC. Q is on BC
7. Draw PS⊥BC. S is on BC
Perform calculation we will get PQ=QR= a.h/(a+h)
And PQRS is a Square