Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Tuesday, November 13, 2018

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

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Let KN & LP meet at O

ReplyDeleteLet 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

Very nice solution Sumith

DeleteThank u Pradyumna

ReplyDeleteIt 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.

ReplyDeletehttps://photos.app.goo.gl/S4Um1QuKJ9PB7ua48

ReplyDeleteLet 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