Let ABCD be a square and P be a point on the side CD (P=C and P=D ). Let AQ and BR be heights in the triangle ABP, and let S be the interesection point of lines CQ and DR. Prove that ∠ASB=90°.
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I will give you some hints.
1. Suppose line AQ intersect side BC at M, and line BR intersect side AD at N. Then QMCP and RPDN are cyclic.
2. In the picture there are many pairs of congruent triangles.
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
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, this inspired me to drop pedals from C,D to PB,PA respectively, denote them X,Y, now we just have to prove △CQX∼△RDY⟺QXCX=DYRY. Now pay some attention to congruent△BCX,△ABQ and △ADX,△BAR and the rest is not hard and can be proceeded multiple ways.
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Top NewestI will give you some hints. 1. Suppose line AQ intersect side BC at M, and line BR intersect side AD at N. Then QMCP and RPDN are cyclic. 2. In the picture there are many pairs of congruent triangles.
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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
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The answer is 392.
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Will not trouble you again
Good luck with the Olympiad training
Des O Carroll
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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
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Sir what if we draw a circle and we use tangent properties to prove the problem
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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, this inspired me to drop pedals from C,D to PB,PA respectively, denote them X,Y, now we just have to prove △CQX∼△RDY⟺QXCX=DYRY. Now pay some attention to congruent△BCX,△ABQ and △ADX,△BAR and the rest is not hard and can be proceeded multiple ways.
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in triangle abc angle a+b=90°, c=?
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