A square and some other points

Let ABCDABCD be a square and PP be a point on the side CDCD (PCP\neq C and PDP\neq D ). Let AQAQ and BRBR be heights in the triangle ABPABP, and let SS be the interesection point of lines CQCQ and DRDR. Prove that ASB=90°\angle ASB=90°.

Note by Jorge Tipe
5 years, 9 months ago

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I will give you some hints. 1. Suppose line AQAQ intersect side BCBC at MM, and line BRBR intersect side ADAD at NN. Then QMCPQMCP and RPDNRPDN are cyclic. 2. In the picture there are many pairs of congruent triangles.

Jorge Tipe - 5 years, 9 months ago

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Thanks Jorge I know about the congruent triangles and I think I can proceed from your first hint to prove the statement in the problem

I am still struggling to find a solution to the least value of z satisfying z^3=a^4+b^4+(a+b)^4 for distinct positive integers a and b

All I want is the answer not the solution

Can you help me

Des O Carroll - 5 years, 9 months ago

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The answer is 392.

Jorge Tipe - 5 years, 9 months ago

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@Jorge Tipe Many thanks Jorge

Will not trouble you again

Good luck with the Olympiad training

Des O Carroll

Des O Carroll - 5 years, 9 months ago

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presumably we can let p be the midpoint of cd and use coordinate geometry

the calculations turn out to be a little awkward as is usual when using this method but it is the only way I can prove the result would love to see other solutions

Des O Carroll - 5 years, 9 months ago

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Sir what if we draw a circle and we use tangent properties to prove the problem

Biswajit Barik - 2 years, 9 months ago

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Jorge's hint was what I first thought of to trigger some cyclics , however I will present another proof(more or less the same but probably differ in motivation since one is more of the wishful thinking type and one is more direct and logical)

It suffices to prove CQP+PRD=90\angle CQP+\angle PRD=90, this inspired me to drop pedals from C,DC,D to PB,PAPB,PA respectively, denote them X,YX,Y, now we just have to prove CQXRDY    CXQX=RYDY\triangle CQX\sim \triangle RDY\iff \frac {CX}{QX}=\frac {RY}{DY}. Now pay some attention to congruentBCX,ABQ\triangle BCX,\triangle ABQ and ADX,BAR\triangle ADX, \triangle BAR and the rest is not hard and can be proceeded multiple ways.

Xuming Liang - 5 years, 8 months ago

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in triangle abc angle a+b=90°, c=?

gauri shankar - 5 years ago

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