Triangle Ratio From A 1952 AHSME

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Introduction

Geometry is a fundamental branch of mathematics that deals with the study of shapes, sizes, and positions of objects. In this article, we will explore a classic problem from the 1952 American High School Mathematics Examination (AHSME) that involves the concept of triangle ratios. The problem is a great example of how geometry can be used to solve problems in a creative and logical way.

The Problem

The problem states that in the figure, CD\overline{CD}, AE\overline{AE}, and BF\overline{BF} are one-third of their respective sides. We are asked to show that AN2:N2N1:N1D=3:3:1\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1. To solve this problem, we need to use our knowledge of geometry and trigonometry to find the lengths of the segments and then use the concept of similar triangles to establish the required ratio.

Understanding the Figure

Before we dive into the solution, let's take a closer look at the figure. We have a triangle ABCABC with points DD, EE, and FF on the sides ABAB, ACAC, and BCBC respectively. The segments CD\overline{CD}, AE\overline{AE}, and BF\overline{BF} are one-third of their respective sides. We are also given points N1N_1 and N2N_2 on the segments AD\overline{AD} and AN2\overline{AN_2} respectively.

Solution

To solve this problem, we need to use the concept of similar triangles. We can start by drawing the altitude CH\overline{CH} from point CC to side ABAB. This will help us to create two similar triangles, CHA\triangle CHA and CHB\triangle CHB. We can then use the fact that the ratio of the corresponding sides of similar triangles is equal to establish the required ratio.

Let's start by finding the length of the segment AN2\overline{AN_2}. We can use the fact that the ratio of the corresponding sides of similar triangles is equal to establish the following equation:

AN2AH=34\frac{AN_2}{AH} = \frac{3}{4}

We can then use the fact that the length of the segment AH\overline{AH} is equal to the length of the segment AB\overline{AB} minus the length of the segment AD\overline{AD} to establish the following equation:

AH=ABADAH = AB - AD

We can then substitute the value of AHAH into the first equation to get:

AN2ABAD=34\frac{AN_2}{AB - AD} = \frac{3}{4}

We can then use the fact that the length of the segment AD\overline{AD} is equal to one-third of the length of the segment AB\overline{AB} to establish the following equation:

AD=13ABAD = \frac{1}{3}AB

We can then substitute the value of ADAD into the previous equation to get:

AN2AB13AB=34\frac{AN_2}{AB - \frac{1}{3}AB} = \frac{3}{4}

We can then simplify the equation to get:

AN223AB=34\frac{AN_2}{\frac{2}{3}AB} = \frac{3}{4}

We can then cross-multiply to get:

4AN2=3×23AB4AN_2 = 3 \times \frac{2}{3}AB

We can then simplify the equation to get:

4AN2=2AB4AN_2 = 2AB

We can then divide both sides of the equation by 4 to get:

AN2=12ABAN_2 = \frac{1}{2}AB

We can then use the fact that the length of the segment AB\overline{AB} is equal to the sum of the lengths of the segments AD\overline{AD} and DB\overline{DB} to establish the following equation:

AB=AD+DBAB = AD + DB

We can then substitute the value of ABAB into the previous equation to get:

12AB=12(AD+DB)\frac{1}{2}AB = \frac{1}{2}(AD + DB)

We can then simplify the equation to get:

AN2=12(AD+DB)AN_2 = \frac{1}{2}(AD + DB)

We can then use the fact that the length of the segment DB\overline{DB} is equal to the length of the segment DC\overline{DC} plus the length of the segment CB\overline{CB} to establish the following equation:

DB=DC+CBDB = DC + CB

We can then substitute the value of DBDB into the previous equation to get:

AN2=12(AD+DC+CB)AN_2 = \frac{1}{2}(AD + DC + CB)

We can then simplify the equation to get:

AN2=12(AD+DC+CB)AN_2 = \frac{1}{2}(AD + DC + CB)

We can then use the fact that the length of the segment DC\overline{DC} is equal to one-third of the length of the segment AC\overline{AC} to establish the following equation:

DC=13ACDC = \frac{1}{3}AC

We can then substitute the value of DCDC into the previous equation to get:

AN2=12(AD+13AC+CB)AN_2 = \frac{1}{2}(AD + \frac{1}{3}AC + CB)

We can then simplify the equation to get:

AN2=12(AD+13AC+CB)AN_2 = \frac{1}{2}(AD + \frac{1}{3}AC + CB)

We can then use the fact that the length of the segment CB\overline{CB} is equal to the length of the segment CA\overline{CA} minus the length of the segment AB\overline{AB} to establish the following equation:

CB=CAABCB = CA - AB

We can then substitute the value of CBCB into the previous equation to get:

AN2=12(AD+13AC+CAAB)AN_2 = \frac{1}{2}(AD + \frac{1}{3}AC + CA - AB)

We can then simplify the equation to get:

AN2=12(AD+43ACAB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AC - AB)

We can then use the fact that the length of the segment AC\overline{AC} is equal to the length of the segment AB\overline{AB} plus the length of the segment BC\overline{BC} to establish the following equation:

AC=AB+BCAC = AB + BC

We can then substitute the value of ACAC into the previous equation to get:

AN2=12(AD+43(AB+BC)AB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}(AB + BC) - AB)

We can then simplify the equation to get:

AN2=12(AD+43AB+43BCAB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AB + \frac{4}{3}BC - AB)

We can then use the fact that the length of the segment BC\overline{BC} is equal to the length of the segment BD\overline{BD} plus the length of the segment DC\overline{DC} to establish the following equation:

BC=BD+DCBC = BD + DC

We can then substitute the value of BCBC into the previous equation to get:

AN2=12(AD+43AB+43(BD+DC)AB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AB + \frac{4}{3}(BD + DC) - AB)

We can then simplify the equation to get:

AN2=12(AD+43AB+43BD+43DCAB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AB + \frac{4}{3}BD + \frac{4}{3}DC - AB)

We can then use the fact that the length of the segment BD\overline{BD} is equal to the length of the segment BA\overline{BA} minus the length of the segment AD\overline{AD} to establish the following equation:

BD=BAADBD = BA - AD

We can then substitute the value of BDBD into the previous equation to get:

AN2=12(AD+43AB+43(BAAD)+43DCAB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AB + \frac{4}{3}(BA - AD) + \frac{4}{3}DC - AB)

We can then simplify the equation to get:

AN2=12(AD+43AB+43BA43AD+43DCAB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AB + \frac{4}{3}BA - \frac{4}{3}AD + \frac{4}{3}DC - AB)

We can then use the fact that the length of the segment BA\overline{BA} is equal to the length of the segment AB\overline{AB} plus the length of the segment AC\overline{AC} to establish the following equation:

BA=AB+ACBA = AB + AC

We can then substitute the value of BABA into the previous equation to get:

AN2=12(AD+43AB+43(AB+AC)43AD+43DCAB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AB + \frac{4}{3}(AB + AC) - \frac{4}{3}AD + \frac{4}{3}DC - AB)

We can then simplify the equation to get:

AN2=12(AD+43AB+43AB+43AC43AD+43DCAB)AN_2 = \frac{1}{2}(AD + \frac{4}{3}AB + \frac{4}{3}AB + \frac{4}{3}AC - \frac{4}{3}AD + \frac{4}{3}DC - AB)

Q: What is the triangle ratio problem from the 1952 AHSME?

A: The triangle ratio problem from the 1952 AHSME is a classic geometry problem that involves finding the ratio of the lengths of three segments in a triangle. The problem states that in the figure, CD\overline{CD}, AE\overline{AE}, and BF\overline{BF} are one-third of their respective sides, and we need to show that AN2:N2N1:N1D=3:3:1\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1.

Q: What is the significance of the triangle ratio problem?

A: The triangle ratio problem is significant because it involves the concept of similar triangles, which is a fundamental concept in geometry. The problem requires the use of similar triangles to establish the required ratio, and it is a great example of how geometry can be used to solve problems in a creative and logical way.

Q: What are the key steps to solving the triangle ratio problem?

A: The key steps to solving the triangle ratio problem are:

  1. Draw the altitude CH\overline{CH} from point CC to side ABAB.
  2. Use the fact that the ratio of the corresponding sides of similar triangles is equal to establish the equation AN2AH=34\frac{AN_2}{AH} = \frac{3}{4}.
  3. Use the fact that the length of the segment AH\overline{AH} is equal to the length of the segment AB\overline{AB} minus the length of the segment AD\overline{AD} to establish the equation AH=ABADAH = AB - AD.
  4. Substitute the value of AHAH into the first equation to get AN2ABAD=34\frac{AN_2}{AB - AD} = \frac{3}{4}.
  5. Use the fact that the length of the segment AD\overline{AD} is equal to one-third of the length of the segment AB\overline{AB} to establish the equation AD=13ABAD = \frac{1}{3}AB.
  6. Substitute the value of ADAD into the previous equation to get AN2AB13AB=34\frac{AN_2}{AB - \frac{1}{3}AB} = \frac{3}{4}.
  7. Simplify the equation to get AN223AB=34\frac{AN_2}{\frac{2}{3}AB} = \frac{3}{4}.
  8. Cross-multiply to get 4AN2=3×23AB4AN_2 = 3 \times \frac{2}{3}AB.
  9. Simplify the equation to get 4AN2=2AB4AN_2 = 2AB.
  10. Divide both sides of the equation by 4 to get AN2=12ABAN_2 = \frac{1}{2}AB.

Q: What is the final answer to the triangle ratio problem?

A: The final answer to the triangle ratio problem is AN2:N2N1:N1D=3:3:1\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1.

Q: What are some common mistakes to avoid when solving the triangle ratio problem?

A: Some common mistakes to avoid when solving the triangle ratio problem are:

  1. Not drawing the altitude CH\overline{CH} from point CC to side ABAB.
  2. Not using the fact that the ratio of the corresponding sides of similar triangles is equal to establish the equation AN2AH=34\frac{AN_2}{AH} = \frac{3}{4}.
  3. Not substituting the value of AHAH into the first equation to get AN2ABAD=34\frac{AN_2}{AB - AD} = \frac{3}{4}.
  4. Not using the fact that the length of the segment AD\overline{AD} is equal to one-third of the length of the segment AB\overline{AB} to establish the equation AD=13ABAD = \frac{1}{3}AB.
  5. Not simplifying the equation to get AN223AB=34\frac{AN_2}{\frac{2}{3}AB} = \frac{3}{4}.

Q: What are some real-world applications of the triangle ratio problem?

A: Some real-world applications of the triangle ratio problem include:

  1. Architecture: The triangle ratio problem can be used to design buildings and bridges.
  2. Engineering: The triangle ratio problem can be used to design machines and mechanisms.
  3. Physics: The triangle ratio problem can be used to describe the motion of objects.
  4. Computer Science: The triangle ratio problem can be used to develop algorithms and data structures.

Q: What are some tips for solving the triangle ratio problem?

A: Some tips for solving the triangle ratio problem are:

  1. Draw a diagram to visualize the problem.
  2. Use the fact that the ratio of the corresponding sides of similar triangles is equal to establish the equation.
  3. Substitute the value of AHAH into the first equation to get AN2ABAD=34\frac{AN_2}{AB - AD} = \frac{3}{4}.
  4. Use the fact that the length of the segment AD\overline{AD} is equal to one-third of the length of the segment AB\overline{AB} to establish the equation AD=13ABAD = \frac{1}{3}AB.
  5. Simplify the equation to get AN223AB=34\frac{AN_2}{\frac{2}{3}AB} = \frac{3}{4}.
  6. Cross-multiply to get 4AN2=3×23AB4AN_2 = 3 \times \frac{2}{3}AB.
  7. Simplify the equation to get 4AN2=2AB4AN_2 = 2AB.
  8. Divide both sides of the equation by 4 to get AN2=12ABAN_2 = \frac{1}{2}AB.

Q: What are some resources for learning more about the triangle ratio problem?

A: Some resources for learning more about the triangle ratio problem are:

  1. Geometry textbooks.
  2. Online resources such as Khan Academy and Mathway.
  3. Video lectures on YouTube.
  4. Practice problems and worksheets.
  5. Online communities and forums.