## Tuesday, November 13, 2018

### Geometry Problem 1400: Triangle with three rectangles on the sides, Perpendicular Bisectors, Concurrency

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Let KN & LP meet at O

Let OB = u, OA = v & OC = w
Let OE = OJ = a, OD = OJ = b, OH = c & OF = d

Using Apollonius for Tr. OBE & ODJ,

2.OP^2 + 2.BP^2 =
a^2 + u^2 = b^2 + v^2 ...(1)

Similarly

b^2 + w^2 = d^2 + u^2 ...(2) and
c^2 + v^2 = a^2 + w^2 ....(3)

Adding (1), (2), (3) gives c = d

So MQ must pass thro O

Sumith Peiris
Moratuwa
Sri Lanka

1. Very nice solution Sumith

3. It is obvious that Triangles BGD, CFH, AJE makes one triangle if the opposite sides of the rectangles (that are not side of the original triangle) are overlapped. In this new triangle, the perpendicular bisectors overlap the KN, LP and MO rays because the sides of the triangle are only pushed away from the intersetion point without changing their direction and length. Because, the perpendicular bisectors of the triangle concur at one point, these rays also concur at one point.

4. https://photos.app.goo.gl/S4Um1QuKJ9PB7ua48

Let S, T and R are circumcenters of triangles HCF, DBG and JAE
We have ∠ (CFH)= ∠ (TSQ)= u1
And ∠ (RSQ)= ∠ (CHF)=v1
Similarly we also have u2, v2, u3, v3 ( see sketch)
Now sin(u1)/sin(v1) x sin(u2)/sin(v2) x sin(u3)/sin(v3)
=CH/CF x AE/AJ x BG/BD= 1
So KN , LP and MQ are concurrent per Ceva’s theorem in triangle STR