Difference between revisions of "2005 AIME I Problems/Problem 15"
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Triangle <math> ABC </math> has <math> BC=20. </math> The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] <math> AD. </math> If the area of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are integers and <math> n </math> is not [[divisor | divisible]] by the [[perfect square | square]] of a prime, find <math> m+n. </math> | Triangle <math> ABC </math> has <math> BC=20. </math> The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] <math> AD. </math> If the area of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are integers and <math> n </math> is not [[divisor | divisible]] by the [[perfect square | square]] of a prime, find <math> m+n. </math> | ||
− | == Solution == | + | == Solution 1== |
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− | Let <math>E</math>, <math>F</math> and <math>G</math> be the points of tangency of the incircle with <math>BC</math>, <math>AC</math> and <math>AB</math>, respectively. Without loss of generality, let <math>AC < AB</math>, so that <math>E</math> is between <math>D</math> and <math>C</math>. Let the length of the median be <math>3m</math>. Then by two applications of the [[Power of a Point Theorem]], <math>DE^2 = 2m \cdot m = AF^2</math>, so <math>DE = AF</math>. Now, <math>CE</math> and <math>CF</math> are two tangents to a circle from the same point, so <math>CE = CF = c</math> and thus <math>AC = AF + CF = DE + CE = CD = 10</math>. Then <math>DE = AF = AG = 10 - c</math> so <math>BG = BE = BD + DE = 20 - c</math> and thus <math>AB = AG + BG = 30 - 2c</math>. | + | Let <math>E</math>, <math>F</math> and <math>G</math> be the points of tangency of the incircle with <math>BC</math>, <math>AC</math> and <math>AB</math>, respectively. Without loss of generality, let <math>AC < AB</math>, so that <math>E</math> is between <math>D</math> and <math>C</math>. Let the length of the median be <math>3m</math>. Then by two applications of the [[Power of a Point Theorem]], <math>DE^2 = 2m \cdot m = AF^2</math>, so <math>DE = AF</math>. Now, <math>CE</math> and <math>CF</math> are two tangents to a circle from the same point, so by the Two Tangent Theorem <math>CE = CF = c</math> and thus <math>AC = AF + CF = DE + CE = CD = 10</math>. Then <math>DE = AF = AG = 10 - c</math> so <math>BG = BE = BD + DE = 20 - c</math> and thus <math>AB = AG + BG = 30 - 2c</math>. |
Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline{AD}</math>, we have | Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline{AD}</math>, we have | ||
− | <cmath>(3m)^2\cdot 20 + 20\cdot10\cdot10 = | + | <cmath>(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.</cmath> |
Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two results to solve for <math>c</math> and we get | Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two results to solve for <math>c</math> and we get | ||
− | <cmath>(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath> | + | <cmath>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath> |
− | Thus <math>c = 2</math> or <math> = 10</math>. We discard the value <math>c = 10</math> as extraneous (it gives us | + | Thus <math>c = 2</math> or <math> = 10</math>. We discard the value <math>c = 10</math> as extraneous (it gives us a line) and are left with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = \boxed{038}</math>. |
+ | |||
+ | == Solution 2 == | ||
+ | WLOG let E be be between C & D (as in solution 1). Assume <math>AD = 3m</math>. We use power of a point to get that | ||
+ | <math>AG = DE = \sqrt{2}m </math> and <math>AB = AG + GB = AG + BE = 10+2\sqrt{2} m</math> | ||
+ | |||
+ | Since now we have <math>AC = 10</math>, <math>BC = 20, AB = 10+2\sqrt{2} m </math> in triangle <math>\triangle ABC</math> and [[cevian]] <math>AD = 3m</math>. Now, we can apply [[Stewart's Theorem]]. | ||
+ | |||
+ | <cmath>2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000</cmath> | ||
+ | <cmath>1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}</cmath> | ||
+ | <cmath>100 m^2 = 400\sqrt{2}m</cmath> | ||
+ | |||
+ | <math>m = 4\sqrt{2}</math> or <math>m = 0</math> if <math>m = 0</math>, we get a degenerate triangle, so <math>m = 4\sqrt{2}</math>, and thus <math>AB = 26</math>. You can now use [[Heron's Formula]] to finish. The answer is <math>24 \sqrt{14}</math>, or <math>\boxed{038}</math>. | ||
+ | |||
+ | -Alexlikemath | ||
+ | |||
+ | == Solution 3 == | ||
+ | Let <math>E, F</math>, and <math>G</math> be the point of tangency (as stated in Solution 1). We can now let <math>AD</math> be <math>3m</math>. By using [[Power of a Point Theorem]] on A to the incircle, you get that <math>AG^2 = 2m^2</math>. We can use it again on point D to the incircle to get the equation <math>(10 - CE)^2 = 2m^2</math>. Setting the two equations equal to each other gives <math>(10 - CE)^2 = AG^2</math>, and it can be further simplified to be <math>10 - CE = AG</math> | ||
+ | |||
+ | Let lengths <math>AC</math> and <math>AB</math> be called <math>b</math> and <math>c</math>, respectively. We can write <math>AG</math> as <math>\frac{b + c - 20}{2}</math> and <math>CE</math> as <math>\frac{b + 20 - c}{2}</math>. Plugging these into the equation, you get: | ||
+ | |||
+ | <cmath>10 - \frac{b + 20 - c}{2} = \frac{b + c - 20}{2}</cmath> | ||
+ | <cmath>b + c - 20 + b + 20 - c = 20 \rightarrow b = 10</cmath> | ||
+ | |||
+ | Additionally, by [[Median of a triangle]] formula, you get that <math>3m = \frac{\sqrt{2c^2 - 200}}{2}</math> | ||
+ | |||
+ | Refer back to the fact that <math>AG^2 = 2m^2</math>. We can now plug in our variables. | ||
+ | |||
+ | <cmath>\left(\frac{b + c - 20}{2}\right)^2 = 2m^2 \rightarrow (c - 10)^2 = 8m^2</cmath> | ||
+ | <cmath>c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}</cmath> | ||
+ | <cmath>9c^2 - 180c + 900 = 4c^2 - 400</cmath> | ||
+ | <cmath>5c^2 - 180c + 1300 = 0</cmath> | ||
+ | <cmath>c^2 - 36c + 260 = 0</cmath> | ||
+ | |||
+ | Solving, you get that <math>c = 26</math> or <math>10</math>, but the latter will result in a degenerate triangle, so <math>c = 26</math>. | ||
+ | Finally, you can use [[Heron's Formula]] to get that the area is <math>24\sqrt{14}</math>, giving an answer of <math>\boxed{038}</math> | ||
+ | |||
+ | ~sky2025 | ||
== See also == | == See also == |
Latest revision as of 18:26, 16 February 2021
Problem
Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find
Solution 1
Let , and be the points of tangency of the incircle with , and , respectively. Without loss of generality, let , so that is between and . Let the length of the median be . Then by two applications of the Power of a Point Theorem, , so . Now, and are two tangents to a circle from the same point, so by the Two Tangent Theorem and thus . Then so and thus .
Now, by Stewart's Theorem in triangle with cevian , we have
Our earlier result from Power of a Point was that , so we combine these two results to solve for and we get
Thus or . We discard the value as extraneous (it gives us a line) and are left with , so our triangle has area and so the answer is .
Solution 2
WLOG let E be be between C & D (as in solution 1). Assume . We use power of a point to get that and
Since now we have , in triangle and cevian . Now, we can apply Stewart's Theorem.
or if , we get a degenerate triangle, so , and thus . You can now use Heron's Formula to finish. The answer is , or .
-Alexlikemath
Solution 3
Let , and be the point of tangency (as stated in Solution 1). We can now let be . By using Power of a Point Theorem on A to the incircle, you get that . We can use it again on point D to the incircle to get the equation . Setting the two equations equal to each other gives , and it can be further simplified to be
Let lengths and be called and , respectively. We can write as and as . Plugging these into the equation, you get:
Additionally, by Median of a triangle formula, you get that
Refer back to the fact that . We can now plug in our variables.
Solving, you get that or , but the latter will result in a degenerate triangle, so . Finally, you can use Heron's Formula to get that the area is , giving an answer of
~sky2025
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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