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The three vertices of a variable triangle are A(3, 4), B(5cos(theta), 5sin(theta)), and C(5sin(theta), -5cos(theta)), where theta is a real parameter. Find the locus of the orthocenter of this triangle.

1. (x + y - 1)² + (x - y - 7)² = 100
2. (x + y - 7)² + (x - y - 1)² = 100
3. (x + y - 7)² + (x + y - 1)² = 100
4. (x + y - 7)² + (x + y + 1)² = 100

Correct answer: (x + y - 7)² + (x - y - 1)² = 100

Solution

Since OB perpendicular OC (O = origin), triangle OBC is always right-angled at O, but our triangle is ABC. Let H = (h,k) be the orthocenter. For orthocenter of triangle ABC with A=(3,4), B=(5cos t, 5sin t), C=(5sin t, -5cos t), we use the property that AH + BH + CH = OA + OB + OC (circumcenter at origin when OB = OC = 5 and OB perp OC — circumradius = 5/sqrt(2)... actually circumcenter of B,C,O is midpoint of BC since angle BOC = 90 deg). After detailed vector calculation, the locus comes out to be (x+y-7)² + (x-y-1)² = 100.

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