Canada Math Olympiad Problem | Best Math Olympiad Problems | Geometry

Math Booster

26 Dec 202318:06

EducationalLearning

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TLDRThe video script presents a geometric problem involving a semicircle inscribed in a right-angled triangle with sides BC=4 and AC=3. It introduces a method to find the length of side 'a' using the properties of a circle, tangents, and Pythagorean theorem. The solution involves calculating the radius 'r', applying it to various parts of the triangle, and solving a quadratic equation to determine 'a'. The script concludes with the correct value of 'a' as 3 times 7, which simplifies to 21.

Takeaways

📐 The problem involves a semicircle inscribed in a right-angled triangle with sides BC and AC given as 4 and 3, respectively.
🔶 The center of the semicircle is denoted as O, which is also the center of the circumscribed circle of triangle ODCE.
⊥ The radii of the semicircle, OE and OD, are perpendicular to the tangent at the points of tangency, creating right angles.
🔴 Since OE = OD = OC = r, triangle ODC is a square, and all sides are equal to the radius r.
✂️ The length of BD is found by subtracting the radius from BC, so BD = 4 - r.
📏 The length of AE is found by subtracting the radius from AC, so AE = 3 - r.
🔄 The area of triangle ABC is expressed as the sum of areas of triangles AOE, ODC, and OBD.
📐 The area of triangle ODC is a square with side length r, so the area is r^2.
📈 The equation for the area of triangle ABC is set up with the given side lengths and the expressions for AE and BD in terms of r.
🧩 After simplifying, the equation 6 = 3r - r^2 + 2r^2 + 4r - 2r^2 is derived, leading to a quadratic equation in r.
🔢 Solving the quadratic equation gives the value of r as 12/7.
📐 Using the Pythagorean theorem on triangle AOE, an equation involving x (the unknown side length), r, and AE is formed.
🔍 Substituting the value of r into the equation and simplifying leads to a quadratic equation in x.
📉 The quadratic formula is applied to find the value of x, resulting in two potential solutions, but only the positive one is meaningful in this geometric context.
🎯 The final value of x is determined to be 3 times 7, which simplifies the solution process.

Q & A

What is the shape described in the video script?
-The shape described is a semicircle inside a right-angled triangle.
What are the lengths of sides BC and AC in the given triangle?
-The length of side BC is 4 units, and the length of side AC is 3 units.

Outlines

00:00

📐 Understanding the Problem Setup

In this video, we are given a semicircle inside a right-angle triangle where BC is 4, AC is 3, and the radius of the semicircle is X. The center of the semicircle is O. The problem involves understanding the relationships between the different segments and angles, noting that radius and tangents are perpendicular to each other, forming a square with segments OD and CE both equal to the radius R. The goal is to find R by setting up and solving various geometric relationships and equations involving these segments.

05:01

📏 Forming and Solving Equations

We continue by expressing BC and AC in terms of R, leading to the equations BD = 4 - R and AE = 3 - R. The area of triangle ABC is calculated using the formula for the area of a triangle. By breaking down triangle ABC into smaller geometric areas involving OE, OD, and BD, we establish equations that involve R. Solving these equations through substitution and simplification, we find R = 12/7.

10:02

🔍 Applying the Pythagorean Theorem

Next, we focus on the triangle formed by AO, OE, and AE, where AO = R + X. Using the Pythagorean theorem, we set up the equation (R + X)^2 = (3 - R)^2 + R^2. Substituting R = 12/7, we derive a quadratic equation in terms of X. This equation is then solved using standard algebraic methods, involving cross-multiplication and simplification, to find the value of X.

15:07

📊 Solving the Quadratic Equation

Finally, we solve the quadratic equation 49X^2 + 168X - 81 = 0 by using the quadratic formula. The coefficients are identified as A = 49, B = 168, and C = -81. We find the roots of the equation, simplifying the radical expressions step-by-step, ultimately determining the possible values for X. However, we encounter and discard a negative solution, confirming that X = 3 * 7, which concludes our problem-solving process.

Mindmap

Keywords

💡Semicircle

A semicircle is half of a circle, formed by drawing a straight line through the center of the circle. In the video's context, the semicircle is a key geometric shape that is inscribed within a right-angled triangle, creating a specific geometric configuration that is central to solving the problem presented.

💡Right-Angled Triangle

A right-angled triangle is a triangle that has one angle measuring exactly 90 degrees. In the script, the right-angled triangle is the main figure within which the semicircle is inscribed, and the lengths of its sides are given as part of the problem setup.

💡Radius

The radius of a circle is the distance from the center of the circle to any point on its circumference. In the video, the radius is denoted by 'r' and is used to establish relationships between the sides of the triangle and the lengths of segments within the semicircle.

💡Tangent

A tangent to a circle is a straight line that touches the circle at exactly one point. In the script, the concept of tangency is used to establish that certain lines within the circle are perpendicular to the radius, which is a crucial property for solving the geometric problem.

💡Perpendicular

Perpendicular lines are lines that intersect at a right angle, 90 degrees. The script uses the term to describe the relationship between the radius and the tangent in a circle, which is essential for the geometric reasoning in the problem.

💡Area

The area of a shape refers to the amount of space it occupies in a two-dimensional plane. In the video, the area is used to set up an equation involving the semicircle and the right-angled triangle, which is necessary to find the unknown length 'X'.

💡Pythagorean Theorem

The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. The script applies this theorem to find the relationship between the sides of the triangle and the radius of the semicircle.

💡Equation

An equation is a mathematical statement that asserts the equality of two expressions. In the script, equations are used to express the relationships between the different lengths and areas in the geometric figure, ultimately leading to the solution for 'X'.

💡Quadratic Equation

A quadratic equation is a polynomial equation of degree two, in the form of ax² + bx + c = 0. The script derives a quadratic equation from the geometric relationships and uses it to solve for the unknown variable 'X'.

💡Root

A root of an equation is a value of the variable that makes the equation true. The script discusses finding the roots of the derived quadratic equation to determine the possible values for 'X'.

💡LCM (Least Common Multiple)

The least common multiple of two or more numbers is the smallest number that is a multiple of each of the numbers. In the script, the LCM is used in the process of solving the quadratic equation, specifically when combining terms to simplify the equation.

Highlights

A semicircle is inscribed within a right-angled triangle with given sides BC and AC.

The center of the semicircle is denoted as point O, which is also the center of the right-angled triangle.

The radius of the semicircle is represented as r, and it is equal to both OE and OD.

Tangents from a point to a circle are perpendicular to the radius at the point of tangency.

Triangle ODC is a square because OD and DC are both equal to the radius r.

BD is calculated as the difference between BC and the radius r.

AE is calculated as the difference between AC and the radius r.

The area of triangle ABC is expressed as the sum of areas of triangles AOE, ODC, and OBD.

The area of triangle ODC is a square with side length r.

An equation is formed to solve for the radius r using the areas of the triangles.

The value of r is found to be 12/7 by solving the equation.

The triangle AOE is right-angled with sides AO, OE, and AE.

AO is calculated as the sum of the radius r and the unknown side X.

Pythagoras' theorem is applied to triangle AOE to find a relationship between X and r.

An equation in terms of X is derived and solved using the quadratic formula.

The value of X is found to be 3 times 7, which simplifies to 21.

The negative solution for X is discarded as it is not possible in this context.

The final value of X is confirmed to be 21 after simplification.

Transcripts

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