Hungary Math Olympiad Problem | Best Math Olympiad Problems | Geometry Problem
TLDRThis video script explores a geometric problem involving a circle inscribed within a rectangle, with tangents of varying lengths from points A, B, C, and D. The challenge is to determine the value of 'X'. The solution uses the properties of tangents being perpendicular to the radius at the point of tangency and applies Pythagoras' theorem to set up equations. By comparing and subtracting these equations, the script derives the value of 'X'. The process involves algebraic manipulation and understanding of geometric relationships, concluding with the calculation of 'X' as the square root of 7361, emphasizing the importance of considering only the positive value due to the physical context of length.
Takeaways
- π The problem involves a circle inscribed in a rectangle with tangents from points A, B, C, and D to the circle.
- β The length of the tangents from points A, B, C, and D to the circle are given as 33, 85, x, and 35 respectively.
- β₯ The script explains that the radius and tangent to a circle are perpendicular to each other.
- πΆ The center of the circle is denoted as O, and angles formed by the radius and tangents are 90Β°.
- π The script uses Pythagoras' theorem to express the lengths of segments AO, BO, CO, and DO in terms of the radius (r) and the tangent lengths.
- π By joining points A, B, C, and D to the center O, the script introduces points M and N to further analyze the geometry.
- π The script introduces variables H1 and H2 to represent the lengths of segments AS and SD, respectively.
- π The script sets up equations for the squares of the lengths of segments AO, BO, CO, and DO based on the geometry of the rectangle and the circle.
- π’ The script derives four equations by comparing the lengths of segments and uses them to find the value of x.
- 𧩠By subtracting these equations, the script simplifies the problem to find the value of x².
- π The final step involves calculating xΒ² by adding and subtracting the appropriate terms and then taking the square root to find the positive value of x.
Q & A
What is the main problem presented in the video?
-The main problem is to find the length of the tangent (x) from point C to a circle that is inscribed inside a rectangle, given the lengths of other tangents from points A, B, and D.
What property of a circle is used in the video to solve the problem?
-The property that a radius and a tangent to a circle are perpendicular to each other is used.
How many angles are mentioned to be 90Β° in the video?
-Four angles are mentioned to be 90Β°, which are the angles formed between the tangents and the radii at the points of contact.
What is the significance of point M in the video?
-Point M is used to create a right triangle by joining AO and OM, which helps in applying the Pythagorean theorem.
What is the relationship between AO and OM in the video?
-AO is the length of the tangent from point A, and OM is the radius of the circle.
What are the equations derived for AO, BO, and CO in terms of s and r?
-The equations are AO = β(AO^2) = β(33^2 + r^2), BO = β(BO^2) = β(85^2 + r^2), and CO = β(CO^2) = β(X^2 + r^2).
How does the video relate the rectangle to the circle and the tangents?
-The video uses the rectangle's sides and the perpendicular lines from the center O to the rectangle's sides to form right triangles with the tangents.
What are the variables H1 and H2 used for in the video?
-H1 and H2 are the lengths of the perpendicular segments from the center O to the sides of the rectangle, which are used in the equations for the tangents.
How many equations are formed in the video to solve for x?
-Four equations are formed, one for each tangent from points A, B, C, and D.
What mathematical operation is used to isolate x in the equations?
-Subtraction of equations is used to isolate x and simplify the equations to find its value.
How does the video conclude the value of x?
-By subtracting and simplifying the equations, the video finds that x^2 equals 7361, and thus x is the square root of 7361, which is approximately 86.
Outlines
π Geometric Relationships in a Circle and Rectangle
This paragraph introduces a geometric problem involving a circle inscribed within a rectangle with tangents from points A, B, C, and D to the circle, each with different lengths: 33, 85, x, and 35 respectively. The script explains that tangents to a circle are perpendicular to the radius at the point of tangency, forming right angles. By drawing lines from the center of the circle (O) to the points of tangency and applying the Pythagorean theorem, the script sets up equations to solve for x, the unknown length of the tangent from point C.
π Constructing Perpendiculars and Forming Equations
The second paragraph continues the problem-solving process by drawing perpendicular lines from the center of the circle (O) to the sides of the rectangle, creating additional points P, Q, S, and T. It introduces variables B1, B2, H1, and H2 to represent segments of these perpendiculars and uses these to form equations based on the Pythagorean theorem. The script then equates these equations to the lengths of the tangents, creating a system of equations to solve for x.
π Solving for x by Subtracting Equations
In this paragraph, the script simplifies the problem by subtracting the equations derived from the geometric setup. This process eliminates variables representing the radius of the circle (r) and the segments H1 and H2, leaving equations that relate the squares of the lengths of the sides of the rectangle (B1, B2) to the squares of the tangent lengths (33, 85, x, 35). The script identifies two key equations (5 and 6) that will be used to solve for x.
π’ Calculating the Value of x
The final paragraph concludes the problem by solving for x. It uses the equations derived in the previous paragraph to find that xΒ² equals the difference between the squares of the lengths of the tangents (85 - 33 + 35). The script then performs arithmetic operations to calculate the value of xΒ² as 7225. It explains the process of finding the square root to determine the positive value of x, which is a length and must be positive. The final answer for x is given as the square root of 7225.
Mindmap
Keywords
π‘Circle
π‘Rectangle
π‘Tangent
π‘Perpendicular
π‘Radius
π‘Pythagorean Theorem
π‘Angle
π‘Equation
π‘Length
π‘Square Root
Highlights
A circle is inscribed inside a rectangle with tangent lengths from points A, B, C, and D being 33, 85, x, and 35 respectively.
The radius and tangent of a circle are perpendicular to each other.
Angles formed by the tangents and the radius are all 90Β°.
By joining the center of the circle to the points of tangency, we can use Pythagoras' theorem to find the lengths.
The length AO is expressed as the square root of the sum of the squares of AM and OM, where OM is the radius.
Similar expressions are derived for lengths BO, CO, and DO using the same method.
A rectangle ABCD is considered with a circle inscribed within it, and perpendicular lines are drawn from the center through points P, Q, S, and T.
The lengths of AP, BP, DQ, and QC are denoted as B1, and lengths AS, SD as H1 and H2.
Expressions for lengths AO, BO, CS, and DS are derived using the right-angled triangles formed.
Four equations are established based on the derived expressions for AO, BO, CS, and DS.
By comparing and subtracting these equations, relationships between B1, B2, H1, H2, and the unknown x are found.
Equations 5 and 6 are derived to relate B2 squared minus B1 squared to the known tangent lengths and x squared.
The value of x squared is calculated by adding and subtracting the squares of the known tangent lengths.
The final value of x squared is found to be 7361 after simplification.
The length x is determined to be the positive square root of 7361, as length cannot be negative.
The process involves algebraic manipulation and application of geometric principles.
The solution requires understanding of right-angled triangles and the Pythagorean theorem.
The video demonstrates a methodical approach to solving geometry problems involving circles and tangents.
Transcripts
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