Geometry for College Students
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Let P,Q, and R be the points of tangency of the incircle of 鈭咥BC with sides BC, CA, and AB, respectively. Then lines AP, BQ, and CR:
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are concurrent.
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Gergonne point
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Point of congruency of lines AP, BQ, and CR, where P,Q, and R are the points of tangency of the incircle of 鈭咥BC.
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Ceva's Theorem: Let AP, BQ, and CR be three lines joining the vertices of 鈭咥BC to pints P, Q, and R on the opposite sides. Then these three lines are concurrent if and only if:
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(AR/RB)(BP/PC)(CQ/QA) = 1
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Given distinct points A and B and a positive number 碌, there is exactly ______ point X such that _________________. Also there is at most _________ other point on the line ____ for which this equation holds.
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one AX/XB = 碌 one AB
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Let AP, BQ, and CR be Cevians in 鈭咥BC, where P, Q, and R lie on sides BC, CA, and AB respectively. If these Cevians are concurrent at point T, then:
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AT/AP = [(AR路CQ) + (QA路RB)] / [AR路CQ) + (QA路RB) + (RB路CQ)]
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Let T be the Gergonne point of 鈭咥BC. If coincides with the incenter or the circumcenter or the orthocenter or the centroid of 鈭咥BC, then:
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The triangle must be an equilateral.
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Let U and V be points on sides of AB and AC, respectively, of 鈭咥BC and suppose UV is parallel to BC. Then the intersection of UC and VB:
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lies on the median of AM.
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The lines joining the vertices of a triangle to the points of tangency of the opposite exscribed circles are:
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concurrent.
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Given three concurrent Cevians in a triangle, the the lines obtained by joining the midpoints of the Cevians to the midpoints of the coorresponding sides are:
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concurrent.
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Suppose Cevians AP, BQ, and CR are concurrent at point T so that 鈭咥BC is decomposed into six small triangles. If areas of 鈭咥RT, 鈭咮PT, and 鈭咰QT are equal, then:
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All six small triangles have equal areas.
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General Form of Ceva's Theorem: Let AP, BQ, and CR be Cevians of 鈭咥BC, where points P, Q, and R lie on lines BC, CA, and AB, respectively. Then these Cevians are either concurrent OR parallel if and only if:
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an ODD number of them are interior and (AR/RB)(BP/PC)(CQ/QA) = 1
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In figure 4.5 on p 135, BQ/QC = ___ and XP/PY = ______. And if XP/PY =BQ/QC, then:
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st/ru; [s(t+u)]/[u(r+s)] then both ratios are equal to 1 and in that case XY is parallel to BC.
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If Cevians AP, BQ, and CR are parallel, then:
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the Cevian product is trivial.
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Angular Cevian Product: Suppose that AP, BQ, and CR are Ceviansin 鈭咥BC. Then the corresponding Cevian product is equal to:
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[sin(鈭燗CR)sin(鈭燘AP)sin(鈭燙BQ)]/[sin(鈭燫CB)sin(鈭燩AC)sin(鈭燪BA)] In particular, the three Cevians are concurrent or parallel if and only if an odd number of them are interior and this angular Cevian product is equal to 1.
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Given an arbitrary 鈭咥BC, build three outward-pointing triangles BCU, CAB, and ABW, each sharing a sides with the original triangle. Assume that 鈭燘AW = 鈭燙AV, 鈭燙BU = 鈭燗BW, and 鈭燗CV = 鈭燘CU. And assume that AU, BV, and CW cut across the interior of 鈭咥BC. Then:
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lines AU, BV, and CW are concurrent.
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The pedal triangle of acute 鈭咥BC is 鈭咲EF, and perpendicular AU, BV, and CW are dropped from the vertices of the original triangle to the sides of the pedal triangle. Then lines AU, BV, and CW are:
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concurrent and the point of concurrence is at the circumcenter.
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If Cevians AP, BQ, and CR of 鈭咥BC are concurrent or parallel, then:
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their isogonal Cevians are also concurrent or parallel.
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Isogonal line:
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The line reflected across an angle bisector.
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Symmedians:
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The isogonals of the medians.
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Lemoine point:
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The point of concurrence of the symmedians.
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Isogonal conjugate:
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The point of concurrence of the isogonal lines.
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Let X be a point other than A, B, or C on the circumcircle of 鈭咥BC. Then:
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the isogonal Cevians of AX, BX, and CX are parallel.
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Let X be any point in the interior of 鈭咥BC and let Y be the circumcenter of the triangle whose vertices are the reflections of X in the sides of 鈭咥BC. Then:
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Y is the isogonal conjugate of X with respect to 鈭咥BC.
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Let ABCDEF be a hexagon inscribed in a circle. Then the diagonals AD, BE, and CF are concurrent if and only if:
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(AB/BC)(CD/DE)(EF/FA)=1
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Given and acute angled 鈭咥BC, show that the lines joining A, B, and C to the midpoints of the nearer sides of the pedal triangle are:
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concurrent
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The only point in the interior of 鈭咥BC that is its own isogonal conjugate is:
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the incenter.
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If 鈭咥BC is not a right triangle then the circumcenter and orthocenter are:
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isogonal conjugates.
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If 鈭燙 = 90 in 鈭咥BC, then the Lemoine point of the triangle lies on:
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the altitude from vertex C.
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If the Lemoine point of 鈭咥BC lies on the altitude from vertex C, then:
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either AC = BC or 鈭燙 = 90.