Lesson A great Math Olympiad level Geometry problem
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<H2>A great Math Olympiad level Geometry problem</H2> <H3>Problem 1</H3>In triangle ABC, point X is on side BC such that AX = 13, BX = 13, CX = 5, and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC. <B>Solution</B> I will solve the problem step by step. Make a sketch and follow my reasoning. <pre> Step 1. Triangle BAC is isosceles: angle C is congruent to angle B Law of sine for triangle AXC, written for angle C, says {{{sin(C)/abs(AX)}}} = {{{2R[AXC]}}}, (1) where {{{R[AXC]}}} is the radius of the circle circumscribed about triangle AXC. Law of sine for triangle AXB, written for angle B, says {{{sin(B)/abs(AX)}}} = {{{2R[AXB]}}}, (2) where {{{R[AXB]}}} is the radius of the circle circumscribed about triangle AXB. We are given that {{{R[AXC]}}} is equal to {{{R[AXB]}}}. Therefore, from (1) and (2) {{{sin(C)/abs(AX)}}} = {{{sin(B)/abs(AX)}}}. (3) It implies that sin(C) = sin(B). Since C and B are angles of triangle BAC, it implies that angle C is congruent to angle B. Thus, triangle ABC is isosceles, and side AC is congruent to side AB. Step 1 is complete. Step 2. Triangles BAC and AXB are similar Indeed, both triangles ABC and AXB are isosceles and have common angle B, which is the base angle for both triangles. Thus step 2 is complete. Step 3. Find the lateral sides AB and AC of triangle BAC Let t be the common length of sides AB and AC of triangle BAC (now unknown). In triangle BAC we know the length of its base BC = 13 + 5 = 18. In triangle AXB we know the length of its lateral sides AX = 13 and BX = 13. Since triangles BAC and AXB are similar, there is a proportion for ratios of corresponding sides {{{abs(BC)/abs(AX)}}} = {{{abs(AB)/abs(AX)}}} (it is base : lateral side = base : lateral side). Substituting the values, this proportion takes the form {{{18/t}}} = {{{t/13}}}. (4) From this proportion, t^2 = 18*13; hence, t = {{{sqrt(18*13)}}}. Thus, we found out the lateral sides of triangle BAC: |AB| = |AC| = {{{sqrt(18*13)}}}. Step 3 is complete. Step 4. Find the area of triangle BAC For now, we know that triangle BAC is isosceles: AB= AC = {{{sqrt(18*13)}}} and its base BC has the length of 13+5 = 18 units. Draw perpendicular AD (the altitude) from vertex A to the base BC. It bisects the base BC in two parts of the length {{{18/2}}} = 9 units. The length h of the altitude AD will be (apply the Pythagoras) h^2 = {{{(sqrt(18*13))^2}}} - {{{9^2}}} = 18*13 - 81 = 153; h = {{{sqrt(153)}}}. Now the area of triangle BAC is {{{area[BAC]}}} = {{{(1/2)*abs(BC)*h}}} = {{{(1/2)*18*sqrt(153)}}} = {{{9*sqrt(153)}}}. At this point, the problem is solved in full. <U>ANSWER</U>. The area of triangle BAC is {{{9*sqrt(153)}}}, or about 111.3238519 square units (approximately). </pre> My other additional lessons on miscellaneous Geometry problems in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Find-the-rate-of-moving-of-the-tip-of-a-shadow.lesson>Find the rate of moving of the tip of a shadow</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-radio-transmitter-accessibility-area.lesson>A radio transmitter accessibility area</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Miscellaneous-geometric-problems.lesson>Miscellaneous geometric problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Miscellaneous-problems-on-parallelograms.lesson>Miscellaneous problems on parallelograms</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Remarkable-properties-of-triangles-into-which-diagonals-divide-a-quadrilateral.lesson>Remarkable properties of triangles into which diagonals divide a quadrilateral</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-trapezoid-divided-in-four-triangles-by-its-diagonals.lesson>A trapezoid divided in four triangles by its diagonals</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/A-problem-on-heptagon.lesson>A problem on a regular heptagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/The-area-of-a-regular-octagon.lesson>The area of a regular octagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/The-fraction-of-the-area-of-a-regular-octagon.lesson>The fraction of the area of a regular octagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Try-to-solve-this-nice-Geometry-problem.lesson>Try to solve these nice Geometry problems !</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Find-the-angle-between-sides-of-folded-triangle.lesson>Find the angle between sides of folded triangle</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-problems-on-three-spheres.lesson>A problem on three spheres</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-sphere-placed-in-an-inverted-cone.lesson>A sphere placed in an inverted cone</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/07-An-upper-level-Geometry-problem-on-special-%2815-30-135%29-triangle.lesson>An upper level Geometry problem on special (15°,30°,135°)-triangle</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Nice-geometry-problem-of-a-Math-Olympiad-level.lesson>Nice geometry problem of a Math Olympiad level</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/OVERVIEW-of-my-lessons-on-additional-misc-Geometry-problems.lesson>OVERVIEW of my additional lessons on miscellaneous advanced Geometry problems</A> To navigate over all topics/lessons of the Online Geometry Textbook use this file/link <A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.
