https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Suli&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:08:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_20&diff=792522003 AMC 12B Problems/Problem 202016-07-11T21:20:39Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
Part of the graph of <math>f(x) = ax^3 + bx^2 + cx + d</math> is shown. What is <math>b</math>?<br />
<br />
[[Image:2003_12B_AMC-20.png]]<br />
<br />
<math>\mathrm{(A)}\ -4<br />
\qquad\mathrm{(B)}\ -2<br />
\qquad\mathrm{(C)}\ 0<br />
\qquad\mathrm{(D)}\ 2<br />
\qquad\mathrm{(E)}\ 4</math><br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Since<br />
<cmath>\begin{align*}<br />
-f(-1) = a - b + c - d = 0 = f(1) = a + b + c + d<br />
\end{align*}</cmath><br />
<br />
It follows that <math>b + d = 0</math>. Also, <math>d = f(0) = 2</math>, so <math>b = -2 \Rightarrow \mathrm{(B)}</math>.<br />
<br />
=== Solution 2 ===<br />
Two of the roots of <math>f(x) = 0</math> are <math>\pm 1</math>, and we let the third one be <math>n</math>. Then <br />
<cmath>a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0</cmath><br />
Notice that <math>f(0) = d = an = 2</math>, so <math>b = -an = -2 \Rightarrow \mathrm{(B)}</math>.<br />
<br />
=== Solution 3 ===<br />
Notice that if <math>g(x) = 2 - 2x^2</math>, then <math>f - g</math> vanishes at <math>x = -1, 0, 1</math> and so<br />
<cmath>f(x) - g(x) = ax(x-1)(x+1) = ax^3 - ax</cmath><br />
implies by <math>x^2</math> coefficient, <math>b + 2 = 0, b = -2 \Rightarrow \mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_20&diff=792512003 AMC 12B Problems/Problem 202016-07-11T21:19:44Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Part of the graph of <math>f(x) = ax^3 + bx^2 + cx + d</math> is shown. What is <math>b</math>?<br />
<br />
[[Image:2003_12B_AMC-20.png]]<br />
<br />
<math>\mathrm{(A)}\ -4<br />
\qquad\mathrm{(B)}\ -2<br />
\qquad\mathrm{(C)}\ 0<br />
\qquad\mathrm{(D)}\ 2<br />
\qquad\mathrm{(E)}\ 4</math><br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Since<br />
<cmath>\begin{align*}<br />
-f(-1) = a - b + c - d = 0 = f(1) = a + b + c + d<br />
\end{align*}</cmath><br />
<br />
It follows that <math>b + d = 0</math>. Also, <math>d = f(0) = 2</math>, so <math>b = -2 \Rightarrow \mathrm{(B)}</math>.<br />
<br />
=== Solution 2 ===<br />
Two of the roots of <math>f(x) = 0</math> are <math>\pm 1</math>, and we let the third one be <math>n</math>. Then <br />
<cmath>a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0</cmath><br />
Notice that <math>f(0) = d = an = 2</math>, so <math>b = -an = -2 \Rightarrow \mathrm{(B)}</math>.<br />
<br />
=== Solution 3 ===<br />
Notice that if <math>g(x) = 2 - 2x^2</math>, then <math>f - g</math> vanishes at <math>x = -1, 0, 1</math> and so<br />
<cmath>f(x) - g(x) = ax(x-1)(x+1) = ax^3 - ax</cmath><br />
implies by <math>x^2</math> coefficient, <math>b + 2 = 0, b = -2 \rightarrow \mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_12&diff=773452016 AIME I Problems/Problem 122016-03-04T23:51:52Z<p>Suli: </p>
<hr />
<div>==Problem ==<br />
Find the least positive integer <math>m</math> such that <math>m^2 - m + 11</math> is a product of at least four not necessarily distinct primes. <br />
==Solution==<br />
We claim <math>m = 132</math>. Note <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>.<br />
<br />
Suppose <math>m < 132</math> and <math>m^2 - m + 11 = pqrs</math> for primes <math>p, q, r, s</math>. We easily verify that <math>p, q, r, s \ge 11</math> using quadratic residue argument. But <math>p, q, r, s = 11</math> yields no integer solution for <math>m</math>. Thus <math>pqrs \ge 11^3 \cdot 13</math>. But this requires <math>m \ge 132</math> from solving the quadratic, contradiction. Hence <math>m = 132</math>.<br />
==See Also==<br />
{{AIME box|year=2016|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_15&diff=773432016 AIME I Problems/Problem 152016-03-04T23:48:46Z<p>Suli: </p>
<hr />
<div>==Problem ==<br />
<br />
Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <math>B</math>, respectively, with line <math>AB</math> closer to point <math>X</math> than to <math>Y</math>. Circle <math>\omega</math> passes through <math>A</math> and <math>B</math> intersecting <math>\omega_1</math> again at <math>D \neq A</math> and intersecting <math>\omega_2</math> again at <math>C \neq B</math>. The three points <math>C</math>, <math>Y</math>, <math>D</math> are collinear, <math>XC = 67</math>, <math>XY = 47</math>, and <math>XD = 37</math>. Find <math>AB^2</math>.<br />
<br />
==Solution==<br />
By radical axis theorem <math>AD, XY, BC</math> concur at point <math>E</math>.<br />
<br />
Let <math>AB</math> and <math>EY</math> intersect at <math>S</math>. Note that because <math>AXDY</math> and <math>CYXB</math> are cyclic, by Miquel theorem <math>AXBE</math> are cyclic as well. Thus<br />
<cmath>\angle AEX = \angle ABX = \angle XCB = \angle XYB</cmath>and<br />
<cmath>\angle XEB = \angle XAB = \angle XDA = \angle XYA.</cmath>Thus <math>AY // EB</math> and <math>YB // EA</math> so <math>AEBY</math> is a parallelogram. Hence <math>AS = SB</math> and <math>SE = SY</math>. But notice that <math>DXE</math> and <math>EXC</math> are similar by <math>AA</math> Similarity, so <math>XE^2 = XD \cdot XC = 37 \cdot 67</math>. But<br />
<cmath>XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.</cmath>Hence <math>AB^2 = 47^2 - 37 \cdot 67 = 270.</math><br />
<br />
==See Also==<br />
{{AIME box|year=2016|n=I|num-b=14|after=Last Question}}<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2010_IMO_Problems/Problem_2&diff=728812010 IMO Problems/Problem 22015-11-11T22:50:06Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Given a triangle <math>ABC</math>, with <math>I</math> as its incenter and <math>\Gamma</math> as its circumcircle, <math>AI</math> intersects <math>\Gamma</math> again at <math>D</math>. Let <math>E</math> be a point on arc <math>BDC</math>, and <math>F</math> a point on the segment <math>BC</math>, such that <math>\angle BAF=\angle CAE< \frac12\angle BAC</math>. If <math>G</math> is the midpoint of <math>IF</math>, prove that the intersection of lines <math>EI</math> and <math>DG</math> lies on <math>\Gamma</math>.<br />
<br />
''Authors: Tai Wai Ming and Wang Chongli, Hong Kong''<br />
<br />
== Solution ==<br />
Note that it suffices to prove alternatively that if <math>EI</math> meets the circle again at <math>J</math> and <math>JD</math> meets <math>IF</math> at <math>G</math>, then <math>G</math> is the midpoint of <math>IF</math>. Let <math>JD</math> meet <math>BC</math> at <math>K</math>.<br />
<br />
Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>.<br />
<br />
Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well.<br />
<br />
Key Lemma. Triangles <math>DKI</math> and <math>DIJ</math> are similar.<br />
Proof. Because triangles <math>DKB</math> and <math>DBJ</math> are similar by AA Similarity (for <math>\angle{KBD}</math> and <math>\angle{BJD}</math> both intercept equally sized arcs), we have <math>BD^2 = BK \cdot BJ</math>. But we know that triangle <math>DBI</math> is isosceles (hint: prove <math>\angle{BID} = \angle{IBD}</math>), and so <math>BI^2 = BK \cdot BJ</math>. Hence, by SAS Similarity, triangles <math>DKI</math> and <math>DIJ</math> are similar, as desired.<br />
<br />
Observation 3. As a result, we have <math>\angle{KID} = \angle{IJD} = \angle{DAE} = \angle{FAD}</math>.<br />
<br />
Observation 4. <math>IK // AF</math>.<br />
<br />
Observation 5. If <math>AF</math> and <math>JD</math> intersect at <math>L</math>, then <math>AJLI</math> is cyclic.<br />
<br />
Observation 6. Because <math>\angle{ALI} = \angle{AJE} = \angle{AJC} + \angle{CJE} = \angle{B} + \angle{AEC} = \angle{B} + \angle{BAF} = \angle{AFC}</math>, we have <math>LI // FK</math>.<br />
<br />
Observation 7. <math>LIKF</math> is a parallelogram, so its diagonals bisect each other, so <math>G</math> is the midpoint of <math>FI</math>, as desired.<br />
<br />
== See Also ==<br />
{{IMO box|year=2010|num-b=1|num-a=3}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1982_USAMO_Problems/Problem_1&diff=712541982 USAMO Problems/Problem 12015-07-20T00:25:53Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
In a party with <math>1982</math> persons, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else.<br />
<br />
== Solution ==<br />
We induct on <math>n</math> to prove that in a party with <math>n</math> people, there must be at least <math>(n-3)</math> people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people <math>A, B, C</math>, who do not know each other.)<br />
<br />
Base case: <math>n = 4</math> is obvious.<br />
<br />
Inductive step: Suppose in a party with <math>k</math> people (with <math>k \ge 4</math>), at least <math>(k-3)</math> people know everyone else. Consider a party with <math>(k+1)</math> people. Take <math>k</math> of the people (leaving another person, <math>A</math>, out) and apply the inductive step to conclude that at least <math>(k-3)</math> people know everyone else in the <math>k</math>-person group, <math>G</math>.<br />
<br />
Now suppose that everyone in the group <math>G</math> knows each other. Then take <math>3</math> of these people and <math>A</math> to deduce that <math>A</math> knows a person <math>B \in G</math>, which means <math>B</math> knows everyone else. Then apply the inductive step on the remaining <math>k</math> people (excluding <math>B</math>) to find <math>(k-3)</math> people out of them that know everyone else (including <math>B</math>, of course). Then these <math>(k-3)</math> people and <math>B</math>, which enumerate <math>(k-2)</math> people, know everyone else.<br />
<br />
Suppose that there exist two people <math>B, C \in G</math> who do not know each other. Because <math>k-3 \ge 1</math>, there exist at least one person in <math>G</math>, person <math>D</math>, who knows everyone else in <math>G</math>. Now, take <math>A, B, C, D</math> and observe that because <math>B, C</math> do not know each other, either <math>A</math> or <math>D</math> knows everyone else of <math>A, B, C, D</math> (by the problem condition), so in particular <math>A</math> and <math>D</math> know each other. Then apply the inductive step on the remaining <math>k</math> people (excluding <math>D</math>) to find <math>(k-3)</math> people out of them that know everyone else (including <math>D</math>, of course). Then these <math>(k-3)</math> people and <math>D</math>, which enumerate <math>(k-2)</math> people, know everyone else.<br />
<br />
This completes the inductive step and thus the proof of this stronger result, which easily implies that at least <math>1982 - 3 = \boxed{1979}</math> people know everyone else.<br />
<br />
== See Also ==<br />
{{USAMO box|year=1982|before=First Question|num-a=2}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Olympiad Combinatorics Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1979_USAMO_Problems/Problem_5&diff=707931979 USAMO Problems/Problem 52015-06-17T01:02:19Z<p>Suli: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>A_1,A_2,...,A_{n+1}</math> be distinct subsets of <math>[n]</math> with <math>|A_1|=|A_2|=\cdots =|A_n|=3</math>. Prove that <math>|A_i\cap A_j|=1</math> for some pair <math>\{i,j\}</math>.<br />
<br />
==Solution==<br />
Suppose the problem statement does not hold. It is clear that <math>n \ge 4</math>. Choose the smallest <math>n</math> such that <math>|A_i \cap A_j| \neq 1</math> for each <math>\{i, j\}</math>.<br />
<br />
First, the <math>(n+1)</math> subsets have <math>3(n+1)</math> elements (some repeated) in conjunction. Because there are <math>n</math> elements of <math>[n]</math> total, by the Pigeonhole Principle one element of <math>[n]</math>, say <math>k</math>, is in at least four subsets. Let these subsets be <math>A_1, A_2, A_3, A_4</math>, without loss of generality, and let <math>A_1</math> have elements <math>k, m, n</math>. Then without loss of generality let <math>A_2</math> have elements <math>k, m, p</math>. If set <math>A_3</math> has elements <math>k, n, p</math>, then simple casework shows that it is impossible to create <math>A_4</math> without having <math>A_4</math> intersect one of <math>A_1, A_2, A_3</math> at exactly one element. Thus assume <math>A_3</math> has elements <math>k, m, q</math>. Then <math>A_4</math> has elements <math>k, m, r</math>. Consider each remaining set <math>A_i</math>. Then <math>A_i</math> either contains both <math>k, m</math> or none of them. Because there are <math>(n-2)</math> distinct elements of <math>[n]</math> apart from <math>k, m</math>, at most <math>(n-2)</math> subsets can contain <math>k, m</math>. Then at least 3 subsets do not contain <math>k, m</math>, and it is easy to see that they are disjoint from those subsets that do contain <math>k, m</math>. Thus, we can partition <math>[n]</math> into two subsets, one of which is the union of the <math>t</math> subsets that do contain <math>k, m</math>, and the other is the union of the <math>(n+1)-t</math> subsets that do not contain <math>k, m</math>. Because the latter subset has <math>(n-t-2) < (n+1)-t - 1</math> elements, we may use infinite descent to contradict the minimality of <math>n</math>. The proof is complete.<br />
==See Also==<br />
{{USAMO box|year=1979|num-b=4|after=Last Question}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Olympiad Combinatorics Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1964_IMO_Problems/Problem_2&diff=707911964 IMO Problems/Problem 22015-06-16T23:45:26Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>a, b, c</math> are the sides of a triangle. Prove that <br />
<br />
<cmath>a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.</cmath><br />
<br />
== Solution ==<br />
We can use the substitution <math>a=x+y</math>, <math>b=x+z</math>, and <math>c=y+z</math> to get<br />
<br />
<cmath>2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)</cmath><br />
<br />
<math>2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz</math><br />
<br />
<cmath>x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz</cmath><br />
<br />
<cmath>\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\geq xyz=\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}</cmath><br />
<br />
This is true by AM-GM. We can work backwards to get that the original inequality is true.<br />
<br />
==Solution 2==<br />
Rearrange to get<br />
<cmath>a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,</cmath><br />
which is true by Schur's inequality.</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_6&diff=703872013 USAMO Problems/Problem 62015-05-21T02:54:53Z<p>Suli: </p>
<hr />
<div>==Problem==<br />
Let <math>ABC</math> be a triangle. Find all points <math>P</math> on segment <math>BC</math> satisfying the following property: If <math>X</math> and <math>Y</math> are the intersections of line <math>PA</math> with the common external tangent lines of the circumcircles of triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.</cmath><br />
<br />
==Solution==<br />
Let circle <math>PAB</math> (i.e. the circumcircle of <math>PAB</math>), <math>PAC</math> be <math>\omega_1, \omega_2</math> with radii <math>r_1</math>, <math>r_2</math> and centers <math>O_1, O_2</math>, respectively, and <math>d</math> be the distance between their centers.<br />
<br />
'''Lemma.''' <math>XY = \frac{r_1 + r_2}{d} \sqrt{d^2 - (r_1 - r_2)^2}.</math><br />
<br />
Proof. Let the external tangent containing <math>X</math> meet <math>\omega_1</math> at <math>X_1</math> and <math>\omega_2</math> at <math>X_2</math>, and let the external tangent containing <math>Y</math> meet <math>\omega_1</math> at <math>Y_1</math> and <math>\omega_2</math> at <math>Y_2</math>. Then clearly <math>X_1 Y_1</math> and <math>X_2 Y_2</math> are parallel (for they are both perpendicular <math>O_1 O_2</math>), and so <math>X_1 Y_1 Y_2 X_2</math> is a trapezoid.<br />
<br />
Now, <math>X_1 X^2 = XA \cdot XP = X_2 X^2</math> by Power of a Point, and so <math>X</math> is the midpoint of <math>X_1 X_2</math>. Similarly, <math>Y</math> is the midpoint of <math>Y_1 Y_2</math>. Hence, <math>XY = \frac{1}{2} (X_1 Y_1 + X_2 Y_2).</math> Let <math>X_1 Y_1</math>, <math>X_2 Y_2</math> meet <math>O_1 O_2</math> s at <math>Z_1, Z_2</math>, respectively. Then by similar triangles and the Pythagorean Theorem we deduce that <math>X_1 Z_1 = \frac{r_1 \sqrt{d^2 - (r_1 - r_2)^2}}{d}</math> and <math>\frac{r_2 \sqrt{d^2 - (r_1 - r_2)^2}}{d}</math>. But it is clear that <math>Z_1</math>, <math>Z_2</math> is the midpoint of <math>X_1 Y_1</math>, <math>X_2 Y_2</math>, respectively, so <math>XY = \frac{(r_1 + r_2)}{d} \sqrt{d^2 - (r_1 - r_2)^2},</math> as desired.<br />
<br />
Lemma 2. Triangles <math>O_1 A O_2</math> and <math>BAC</math> are similar.<br />
<br />
Proof. <math>\angle{AO_1 O_2} = \frac{\angle{PO_1 A}}{2} = \angle{ABC}</math> and similarly <math>\angle{AO_2 O_1} = \angle{ACB}</math>, so the triangles are similar by AA Similarity.<br />
<br />
Also, let <math>O_1 O_2</math> intersect <math>AP</math> at <math>Z</math>. Then obviously <math>Z</math> is the midpoint of <math>AP</math> and <math>AZ</math> is an altitude of triangle <math>A O_1 O_2</math>.Thus, we can simplify our expression of <math>XY</math>:<br />
<cmath>XY = \frac{AB + AC}{BC} \cdot \frac{AP}{2 h_a} \sqrt{BC^2 - (AB - AC)^2},</cmath><br />
where <math>h_a</math> is the length of the altitude from <math>A</math> in triangle <math>ABC</math>. Hence, substituting into our condition and using <math>AB = c, BC = a, CA = b</math> gives<br />
<cmath>\left( \frac{2a h_a}{(b+c) \sqrt{a^2 - (b-c)^2}} \right)^2 + \frac{PB \cdot PC}{bc} = 1.</cmath><br />
Using <math>2 a h_a = 4[ABC] = \sqrt{(a + b + c)(a + b - c)(a - b + c)(-a + b + c)}</math> by Heron's Formula (where <math>[ABC]</math> is the area of triangle <math>ABC</math>, our condition becomes<br />
<cmath>\frac{(a + b + c)(-a + b + c)}{(b + c)^2} + \frac{PB \cdot PC}{bc} = 1,</cmath><br />
which by <math>(a + b + c)(-a + b + c) = (b + c)^2 - a^2</math> becomes<br />
<cmath>\frac{PB \cdot PC}{bc} = \frac{a^2 bc}{(b+c)^2}.</cmath><br />
Let <math>PB = x</math>; then <math>PC = a - x</math>. The quadratic in <math>x</math> is<br />
<cmath>x^2 - ax + \frac{a^2 bc}{(b+c)^2} = 0,</cmath><br />
which factors as<br />
<cmath>(x - \frac{ab}{b+c})(x - \frac{ac}{b+c}) = 0.</cmath><br />
Hence, <math>PB = \frac{ab}{b+c}</math> or <math>\frac{ac}{b+c}</math>, and so the <math>P</math> corresponding to these lengths are our answer.<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=703862013 USAMO Problems/Problem 12015-05-20T23:26:22Z<p>Suli: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively. Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively. Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
Firstly, it is easy to see by that <math>\omega_A, \omega_B, \omega_C</math> concur at a point <math>M</math>. Let <math>XM</math> meet <math>\omega_B, \omega_C</math> again at <math>D</math> and <math>E</math>, respectively. Then by Power of a Point, we have <cmath>XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM \cdot XD = XY \cdot XP</cmath> Thusly <cmath>\frac{XY}{XZ} = \frac{XD}{XE}</cmath> But we claim that <math>\triangle XDP \sim \triangle PBM</math>. Indeed, <cmath>\measuredangle XDP = \measuredangle MDP = \measuredangle MBP = - \measuredangle PBM</cmath> and <cmath>\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM</cmath><br />
Therefore, <math>\frac{XD}{XP} = \frac{PB}{PM}</math>. Analogously we find that <math>\frac{XE}{XP} = \frac{PC}{PM}</math> and we are done.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=703852013 USAMO Problems/Problem 12015-05-20T23:26:00Z<p>Suli: </p>
<hr />
<div>==Problem==<br />
In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively. Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively. Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
Firstly, it is easy to see by that <math>\omega_A, \omega_B, \omega_C</math> concur at a point <math>M</math>. Let <math>XM</math> meet <math>\omega_B, \omega_C</math> again at <math>D</math> and <math>E</math>, respectively. Then by Power of a Point, we have <cmath>XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM \cdot XD = XY \cdot XP</cmath> Thusly <cmath>\frac{XY}{XZ} = \frac{XD}{XE}</cmath> But we claim that <math>\triangle XDP \sim \triangle PBM</math>. Indeed, <cmath>\measuredangle XDP = \measuredangle MDP = \measuredangle MBP = - \measuredangle PBM</cmath> and <cmath>\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM</cmath><br />
Therefore, <math>\frac{XD}{XP} = \frac{PB}{PM}</math>. Analogously we find that <math>\frac{XE}{XP} = \frac{PC}{PM}</math> and we are done.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2011_USAMO_Problems/Problem_5&diff=703812011 USAMO Problems/Problem 52015-05-20T22:03:14Z<p>Suli: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>P</math> be a given point inside quadrilateral <math>ABCD</math>. Points <math>Q_1</math> and <math>Q_2</math> are located within <math>ABCD</math> such that <math>\angle Q_1 BC = \angle ABP</math>, <math>\angle Q_1 CB = \angle DCP</math>, <math>\angle Q_2 AD = \angle BAP</math>, <math>\angle Q_2 DA = \angle CDP</math>. Prove that <math>\overline{Q_1 Q_2} \parallel \overline{AB}</math> if and only if <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math>.<br />
<br />
==Solution==<br />
First note that <math>\overline{Q_1 Q_2} \parallel \overline{AB}</math> if and only if the altitudes from <math>Q_1</math> and <math>Q_2</math> to <math>\overline{AB}</math> are the same, or <math>|Q_1B|\sin \angle ABQ_1 =|Q_2A|\sin \angle BAQ_2</math>. Similarly <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math> iff <math>|Q_1C|\sin \angle DCQ_1 =|Q_2D|\sin \angle CDQ_2</math>.<br />
<br />
<br />
If we define <math>S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2}\cdot\frac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}</math>, then we are done if we can show that S=1.<br />
<br />
<br />
By the law of sines, <math>\frac{|Q_1B|}{|Q_1C|}=\frac{\sin\angle Q_1CB}{\sin\angle Q_1BC}</math> and <math>\frac{|Q_2D|}{|Q_2A|}=\frac{\sin\angle Q_2AD}{\sin\angle Q_2DA}</math>.<br />
<br />
<br />
So, <math>S=\frac{\sin \angle ABQ_1}{\sin \angle BAQ_2}\cdot\frac{\sin \angle CDQ_2}{\sin \angle DCQ_1}\cdot\frac{\sin \angle BCQ_1}{\sin \angle CBQ_1}\cdot\frac{\sin \angle DAQ_2}{\sin \angle ADQ_2}</math><br />
<br />
<br />
By the terms of the problem, <math>S=\frac{\sin \angle PBC}{\sin \angle PAD}\cdot\frac{\sin \angle PDA}{\sin \angle PCB}\cdot\frac{\sin \angle PCD}{\sin \angle PBA}\cdot\frac{\sin \angle PAB}{\sin \angle PDC}</math>. (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.) <br />
<br />
<br />
Rearranging yields <math>S= \frac{\sin \angle PBC}{\sin \angle PCB}\cdot\frac{\sin \angle PDA}{\sin \angle PAD}\cdot\frac{\sin \angle PCD}{\sin \angle PDC}\cdot\frac{\sin \angle PAB}{\sin \angle PBA}</math>.<br />
<br />
<br />
Applying the law of sines to the triangles with vertices at P yields <math>S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1</math>.<br />
<br />
==Solution 2==<br />
Lemma. If <math>AB</math> and <math>CD</math> are not parallel, then <math>AB, CD, Q_1 Q_2</math> are concurrent.<br />
<br />
Proof. Let <math>AB</math> and <math>CD</math> meet at <math>R</math>. Notice that with respect to triangle <math>ADR</math>, <math>P</math> and <math>Q_2</math> are isogonal conjugates. With respect to triangle <math>BCR</math>, <math>P</math> and <math>Q_1</math> are isogonal conjugates. Therefore, <math>Q_1</math> and <math>Q_2</math> lie on the reflection of <math>RP</math> in the angle bisector of <math>\angle{DRA}</math>, so <math>R, Q_1, Q_2</math> are collinear. Hence, <math>AB, CD, Q_1 Q_2</math> are concurrent at <math>R</math>.<br />
<br />
Now suppose <math>Q_1 Q_2 // AB</math> but <math>Q_1 Q_2</math> is not parallel to <math>CD</math>. Then <math>AB</math> and <math>CD</math> are not parallel and thus intersect at a point <math>R</math>. But then <math>Q_1 Q_2</math> also passes through <math>R</math>, contradicting <math>Q_1 Q_2 // AB</math>. A similar contradiction occurs if <math>Q_1 Q_2 // CD</math> but <math>Q_1 Q_2</math> is not parallel to <math>CD</math>, so we can conclude that <math>Q_1 Q_2 // AB</math> if and only if <math>Q_1 Q_2 // CD</math>.<br />
<br />
{{MAA Notice}}<br />
<br />
==See also==<br />
*[[USAMO Problems and Solutions]]<br />
<br />
{{USAMO newbox|year=2011|num-b=4|num-a=6}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_1&diff=703372010 USAMO Problems/Problem 12015-05-20T02:55:33Z<p>Suli: /* Footnote to the Footnote */</p>
<hr />
<div>==Problem==<br />
Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter<br />
<math>AB</math>. Denote by <math>P, Q, R, S</math> the feet of the perpendiculars from <math>Y</math> onto<br />
lines <math>AX, BX, AZ, BZ</math>, respectively. Prove that the acute angle<br />
formed by lines <math>PQ</math> and <math>RS</math> is half the size of <math>\angle XOZ</math>, where<br />
<math>O</math> is the midpoint of segment <math>AB</math>.<br />
<br />
==Solution==<br />
Let <math>\alpha = \angle BAZ</math>, <math>\beta = \angle ABX</math>.<br />
Since <math>XY</math> is a chord of the circle with diameter <math>AB</math>,<br />
<math>\angle XAY = \angle XBY = \gamma</math>. From the chord <math>YZ</math>,<br />
we conclude <math>\angle YAZ = \angle YBZ = \delta</math>.<br />
<center><br />
<asy><br />
import olympiad;<br />
<br />
// Scale<br />
unitsize(1inch);<br />
real r = 1.75;<br />
<br />
// Semi-circle: centre O, radius r, diameter A--B.<br />
pair O = (0,0); dot(O); label("$O$", O, plain.S);<br />
pair A = r * plain.W; dot(A); label("$A$", A, unit(A));<br />
pair B = r * plain.E; dot(B); label("$B$", B, unit(B));<br />
draw(arc(O, r, 0, 180)--cycle);<br />
<br />
// points X, Y, Z<br />
real alpha = 22.5;<br />
real beta = 15;<br />
real delta = 30;<br />
pair X = r * dir(180 - 2*beta); dot(X); label("$X$", X, unit(X));<br />
pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y));<br />
pair Z = r * dir(2*alpha); dot(Z); label("$Z$", Z, unit(Z));<br />
<br />
// Feet of perpendiculars from Y<br />
pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P);<br />
pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q);<br />
pair R = foot(Y, B, Z); dot(R); label("$S$", R, unit(R-Y)); dot(R);<br />
pair S = foot(Y, A, Z); dot(S); label("$R$", S, unit(B-S)); dot(S);<br />
pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T);<br />
<br />
// Segments<br />
draw(B--X); draw(B--Y); draw(B--R);<br />
draw(A--Z); draw(A--Y); draw(A--P);<br />
draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S);<br />
draw(R--T); draw(P--T);<br />
<br />
// Right angles<br />
draw(rightanglemark(A, X, B, 3));<br />
draw(rightanglemark(A, Y, B, 3));<br />
draw(rightanglemark(A, Z, B, 3));<br />
draw(rightanglemark(A, P, Y, 3));<br />
draw(rightanglemark(Y, R, B, 3));<br />
draw(rightanglemark(Y, S, A, 3));<br />
draw(rightanglemark(B, Q, Y, 3));<br />
<br />
// Acute angles<br />
import markers;<br />
void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0)<br />
{<br />
string sl = "$\scriptstyle{" + l + "}$";<br />
marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker;<br />
markangle(Label(sl), radius=r, n=n, A, B, C, m);<br />
}<br />
langle(B, A, Z, "\alpha" );<br />
langle(X, B, A, "\beta", n=2);<br />
langle(Y, A, X, "\gamma", nm=1);<br />
langle(Y, B, X, "\gamma", nm=1);<br />
langle(Z, A, Y, "\delta", nm=2);<br />
langle(Z, B, Y, "\delta", nm=2);<br />
langle(R, S, Y, "\alpha+\delta", r=23);<br />
langle(Y, Q, P, "\beta+\gamma", r=23);<br />
langle(R, T, P, "\chi", r=15);<br />
</asy><br />
</center><br />
<br />
Triangles <math>BQY</math> and <math>APY</math> are both right-triangles, and share the<br />
angle <math>\gamma</math>, therefore they are similar, and so the ratio <math>PY :<br />
YQ = AY : YB</math>. Now by [[Thales' theorem]] the angles <math>\angle AXB =<br />
\angle AYB = \angle AZB</math> are all right-angles. Also, <math>\angle PYQ</math>,<br />
being the fourth angle in a quadrilateral with 3 right-angles is<br />
again a right-angle. Therefore <math>\triangle PYQ \sim \triangle AYB</math> and<br />
<math>\angle YQP = \angle YBA = \gamma + \beta</math>.<br />
Similarly, <math>RY : YS = BY : YA</math>, and so <math>\angle YSR = \angle YAB = \alpha + \delta</math>.<br />
<br />
Now <math>RY</math> is perpendicular to <math>AZ</math> so the direction <math>RY</math> is <math>\alpha</math> counterclockwise from the vertical, and since <math>\angle YRS = \alpha + \delta</math> we see that <math>SR</math> is <math>\delta</math> clockwise from the vertical.<br />
<br />
Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> counterclockwise from the vertical.<br />
<br />
Therefore the lines <math>PQ</math> and <math>RS</math> intersect at an angle <math>\chi = \gamma<br />
+ \delta</math>. Now by the central angle theorem <math>2\gamma = \angle XOY</math><br />
and <math>2\delta = \angle YOZ</math>, and so <math>2(\gamma + \delta) = \angle XOZ</math>,<br />
and we are done.<br />
<br />
''Note that <math>RTQY</math> is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?''<br />
<br />
===Footnote===<br />
We can prove a bit more. Namely, the extensions of the segments<br />
<math>RS</math> and <math>PQ</math> meet at a point on the diameter <math>AB</math> that is vertically<br />
below the point <math>Y</math>.<br />
<br />
Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise<br />
from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math><br />
horizontally to the right of <math>Y</math>.<br />
<br />
Now <math>AS = AY \cos(\delta)</math>, so <math>S</math> is <math>AS \sin(\alpha) = AY<br />
\cos(\delta)\sin(\alpha)</math> vertically above the diameter <math>AB</math>. Also,<br />
the segment <math>SR</math> is inclined <math>\delta</math> clockwise from the vertical,<br />
so if we extend it down from <math>S</math> towards the diameter <math>AB</math> it will<br />
meet the diameter at a point which is<br />
<math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math><br />
horizontally to the left of <math>S</math>. This places the intersection point<br />
of <math>RS</math> and <math>AB</math> vertically below <math>Y</math>.<br />
<br />
Similarly, and by symmetry the intersection point of <math>PQ</math> and <math>AB</math><br />
is directly below <math>Y</math> on <math>AB</math>, so the lines through <math>PQ</math> and <math>RS</math><br />
meet at a point <math>T</math> on the diameter that is vertically below <math>Y</math>.<br />
<br />
===Footnote to the Footnote===<br />
The Footnote's claim is more easily proved as follows.<br />
<br />
Note that because <math>\angle{QPY}</math> and <math>\angle{YAB}</math> are both complementary to <math>\beta + \gamma</math>, they must be equal. Now, let <math>PQ</math> intersect diameter <math>AB</math> at <math>T'</math>. Then <math>PYT'A</math> is cyclic and so <math>\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ</math>. Hence <math>T'YSB</math> is cyclic as well, and so we deduce that <math>\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.</math> Hence <math>S, R, T'</math> are collinear and so <math>T = T'</math>. This proves the Footnote.<br />
<br />
== See Also ==<br />
{{USAMO newbox|year=2010|before=First problem|num-a=2}}<br />
{{USAJMO newbox|year=2010|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_1&diff=703362010 USAMO Problems/Problem 12015-05-20T02:55:09Z<p>Suli: </p>
<hr />
<div>==Problem==<br />
Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter<br />
<math>AB</math>. Denote by <math>P, Q, R, S</math> the feet of the perpendiculars from <math>Y</math> onto<br />
lines <math>AX, BX, AZ, BZ</math>, respectively. Prove that the acute angle<br />
formed by lines <math>PQ</math> and <math>RS</math> is half the size of <math>\angle XOZ</math>, where<br />
<math>O</math> is the midpoint of segment <math>AB</math>.<br />
<br />
==Solution==<br />
Let <math>\alpha = \angle BAZ</math>, <math>\beta = \angle ABX</math>.<br />
Since <math>XY</math> is a chord of the circle with diameter <math>AB</math>,<br />
<math>\angle XAY = \angle XBY = \gamma</math>. From the chord <math>YZ</math>,<br />
we conclude <math>\angle YAZ = \angle YBZ = \delta</math>.<br />
<center><br />
<asy><br />
import olympiad;<br />
<br />
// Scale<br />
unitsize(1inch);<br />
real r = 1.75;<br />
<br />
// Semi-circle: centre O, radius r, diameter A--B.<br />
pair O = (0,0); dot(O); label("$O$", O, plain.S);<br />
pair A = r * plain.W; dot(A); label("$A$", A, unit(A));<br />
pair B = r * plain.E; dot(B); label("$B$", B, unit(B));<br />
draw(arc(O, r, 0, 180)--cycle);<br />
<br />
// points X, Y, Z<br />
real alpha = 22.5;<br />
real beta = 15;<br />
real delta = 30;<br />
pair X = r * dir(180 - 2*beta); dot(X); label("$X$", X, unit(X));<br />
pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y));<br />
pair Z = r * dir(2*alpha); dot(Z); label("$Z$", Z, unit(Z));<br />
<br />
// Feet of perpendiculars from Y<br />
pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P);<br />
pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q);<br />
pair R = foot(Y, B, Z); dot(R); label("$S$", R, unit(R-Y)); dot(R);<br />
pair S = foot(Y, A, Z); dot(S); label("$R$", S, unit(B-S)); dot(S);<br />
pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T);<br />
<br />
// Segments<br />
draw(B--X); draw(B--Y); draw(B--R);<br />
draw(A--Z); draw(A--Y); draw(A--P);<br />
draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S);<br />
draw(R--T); draw(P--T);<br />
<br />
// Right angles<br />
draw(rightanglemark(A, X, B, 3));<br />
draw(rightanglemark(A, Y, B, 3));<br />
draw(rightanglemark(A, Z, B, 3));<br />
draw(rightanglemark(A, P, Y, 3));<br />
draw(rightanglemark(Y, R, B, 3));<br />
draw(rightanglemark(Y, S, A, 3));<br />
draw(rightanglemark(B, Q, Y, 3));<br />
<br />
// Acute angles<br />
import markers;<br />
void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0)<br />
{<br />
string sl = "$\scriptstyle{" + l + "}$";<br />
marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker;<br />
markangle(Label(sl), radius=r, n=n, A, B, C, m);<br />
}<br />
langle(B, A, Z, "\alpha" );<br />
langle(X, B, A, "\beta", n=2);<br />
langle(Y, A, X, "\gamma", nm=1);<br />
langle(Y, B, X, "\gamma", nm=1);<br />
langle(Z, A, Y, "\delta", nm=2);<br />
langle(Z, B, Y, "\delta", nm=2);<br />
langle(R, S, Y, "\alpha+\delta", r=23);<br />
langle(Y, Q, P, "\beta+\gamma", r=23);<br />
langle(R, T, P, "\chi", r=15);<br />
</asy><br />
</center><br />
<br />
Triangles <math>BQY</math> and <math>APY</math> are both right-triangles, and share the<br />
angle <math>\gamma</math>, therefore they are similar, and so the ratio <math>PY :<br />
YQ = AY : YB</math>. Now by [[Thales' theorem]] the angles <math>\angle AXB =<br />
\angle AYB = \angle AZB</math> are all right-angles. Also, <math>\angle PYQ</math>,<br />
being the fourth angle in a quadrilateral with 3 right-angles is<br />
again a right-angle. Therefore <math>\triangle PYQ \sim \triangle AYB</math> and<br />
<math>\angle YQP = \angle YBA = \gamma + \beta</math>.<br />
Similarly, <math>RY : YS = BY : YA</math>, and so <math>\angle YSR = \angle YAB = \alpha + \delta</math>.<br />
<br />
Now <math>RY</math> is perpendicular to <math>AZ</math> so the direction <math>RY</math> is <math>\alpha</math> counterclockwise from the vertical, and since <math>\angle YRS = \alpha + \delta</math> we see that <math>SR</math> is <math>\delta</math> clockwise from the vertical.<br />
<br />
Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> counterclockwise from the vertical.<br />
<br />
Therefore the lines <math>PQ</math> and <math>RS</math> intersect at an angle <math>\chi = \gamma<br />
+ \delta</math>. Now by the central angle theorem <math>2\gamma = \angle XOY</math><br />
and <math>2\delta = \angle YOZ</math>, and so <math>2(\gamma + \delta) = \angle XOZ</math>,<br />
and we are done.<br />
<br />
''Note that <math>RTQY</math> is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?''<br />
<br />
===Footnote===<br />
We can prove a bit more. Namely, the extensions of the segments<br />
<math>RS</math> and <math>PQ</math> meet at a point on the diameter <math>AB</math> that is vertically<br />
below the point <math>Y</math>.<br />
<br />
Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise<br />
from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math><br />
horizontally to the right of <math>Y</math>.<br />
<br />
Now <math>AS = AY \cos(\delta)</math>, so <math>S</math> is <math>AS \sin(\alpha) = AY<br />
\cos(\delta)\sin(\alpha)</math> vertically above the diameter <math>AB</math>. Also,<br />
the segment <math>SR</math> is inclined <math>\delta</math> clockwise from the vertical,<br />
so if we extend it down from <math>S</math> towards the diameter <math>AB</math> it will<br />
meet the diameter at a point which is<br />
<math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math><br />
horizontally to the left of <math>S</math>. This places the intersection point<br />
of <math>RS</math> and <math>AB</math> vertically below <math>Y</math>.<br />
<br />
Similarly, and by symmetry the intersection point of <math>PQ</math> and <math>AB</math><br />
is directly below <math>Y</math> on <math>AB</math>, so the lines through <math>PQ</math> and <math>RS</math><br />
meet at a point <math>T</math> on the diameter that is vertically below <math>Y</math>.<br />
<br />
==Footnote to the Footnote==<br />
The Footnote's claim is more easily proved as follows.<br />
<br />
Note that because <math>\angle{QPY}</math> and <math>\angle{YAB}</math> are both complementary to <math>\beta + \gamma</math>, they must be equal. Now, let <math>PQ</math> intersect diameter <math>AB</math> at <math>T'</math>. Then <math>PYT'A</math> is cyclic and so <math>\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ</math>. Hence <math>T'YSB</math> is cyclic as well, and so we deduce that <math>\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.</math> Hence <math>S, R, T'</math> are collinear and so <math>T = T'</math>. This proves the Footnote.<br />
<br />
== See Also ==<br />
{{USAMO newbox|year=2010|before=First problem|num-a=2}}<br />
{{USAJMO newbox|year=2010|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1977_USAMO_Problems/Problem_2&diff=703321977 USAMO Problems/Problem 22015-05-20T02:24:10Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
<math> ABC</math> and <math> A'B'C'</math> are two triangles in the same plane such that the lines <math> AA',BB',CC'</math> are mutually parallel. Let <math> [ABC]</math> denote the area of triangle <math> ABC</math> with an appropriate <math> \pm</math> sign, etc.; prove that<br />
<cmath> 3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB].</cmath><br />
<br />
== Hint ==<br />
Let the parallel lines <math>AA', BB', CC'</math> be parallel to the <math>x-axis</math>, and choose arbitrary origin. Then we can define <math>A(x_1, a), A'(x_2, a), B(y_1, b), B'(y_2, b), C(z_1, c), C'(z_2, c),</math> and so, by the area of a triangle formula, it suffices to prove an algebraic statement that is readily shown to be true.<br />
<br />
== See Also ==<br />
{{USAMO box|year=1977|num-b=1|num-a=3}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1970_IMO_Problems/Problem_1&diff=703311970 IMO Problems/Problem 12015-05-20T02:12:40Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Let <math>M</math> be a point on the side <math>AB</math> of <math>\triangle ABC</math>. Let <math>r_1, r_2</math>, and <math>r</math> be the inscribed circles of triangles <math>AMC, BMC</math>, and <math>ABC</math>. Let <math>q_1, q_2</math>, and <math>q</math> be the radii of the exscribed circles of the same triangles that lie in the angle <math>ACB</math>. Prove that<br />
<br />
<center><br />
<math>\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}</math>.<br />
</center><br />
<br />
== Solution ==<br />
<br />
We use the conventional triangle notations.<br />
<br />
Let <math>I</math> be the incenter of <math>ABC</math>, and let <math>I_{c}</math> be its excenter to side <math>c</math>. We observe that<br />
<br />
<center><br />
<math> r \left[ \cot\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right) \right] = c</math>,<br />
</center><br />
<br />
and likewise,<br />
<br />
<center><br />
<math> \begin{matrix}<br />
c & = &q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\\<br />
& = &q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}</math><br />
</center><br />
<br />
Simplifying the quotient of these expressions, we obtain the result<br />
<br />
<center><br />
<math>\frac{r}{q} = \tan (A/2) \tan (B/2)</math>.<br />
</center><br />
<br />
Thus we wish to prove that<br />
<br />
<center><br />
<math>\tan (A/2) \tan (B/2) = \tan (A/2) \tan (AMC/2) \tan (B/2) \tan (CMB/2)</math>.<br />
</center><br />
<br />
But this follows from the fact that the angles <math>AMC</math> and <math>CBM</math> are supplementary.<br />
<br />
==Solution 2==<br />
By similar triangles and the fact that both centers lie on the angle bisector of <math>\angle{C}</math>, we have <math>\frac{r}{q} = \frac{s-c}{s} = \frac{a + b - c}{a + b + c}</math>, where <math>s</math> is the semi-perimeter of <math>ABC</math>. Let <math>ABC</math> have sides <math>a, b, c</math>, and let <math>AM = c_1, MB = c_2, MC = d</math>. After simple computations, we see that the condition, whose equivalent form is<br />
<cmath>\frac{b + d - c_1}{b + d + c_1} \cdot \frac{a + d - c_2}{a + d + c_2} = \frac{a + b - c}{a + b + c},</cmath><br />
is also equivalent to Stewart's Theorem<br />
<cmath>d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.</cmath><br />
{{alternate solutions}}<br />
<br />
{{IMO box|year=1970|before=First question|num-a=2}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1970_IMO_Problems/Problem_1&diff=703301970 IMO Problems/Problem 12015-05-20T02:12:17Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>M</math> be a point on the side <math>AB</math> of <math>\triangle ABC</math>. Let <math>r_1, r_2</math>, and <math>r</math> be the inscribed circles of triangles <math>AMC, BMC</math>, and <math>ABC</math>. Let <math>q_1, q_2</math>, and <math>q</math> be the radii of the exscribed circles of the same triangles that lie in the angle <math>ACB</math>. Prove that<br />
<br />
<center><br />
<math>\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}</math>.<br />
</center><br />
<br />
== Solution ==<br />
<br />
We use the conventional triangle notations.<br />
<br />
Let <math>I</math> be the incenter of <math>ABC</math>, and let <math>I_{c}</math> be its excenter to side <math>c</math>. We observe that<br />
<br />
<center><br />
<math> r \left[ \cot\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right) \right] = c</math>,<br />
</center><br />
<br />
and likewise,<br />
<br />
<center><br />
<math> \begin{matrix}<br />
c & = &q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\\<br />
& = &q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}</math><br />
</center><br />
<br />
Simplifying the quotient of these expressions, we obtain the result<br />
<br />
<center><br />
<math>\frac{r}{q} = \tan (A/2) \tan (B/2)</math>.<br />
</center><br />
<br />
Thus we wish to prove that<br />
<br />
<center><br />
<math>\tan (A/2) \tan (B/2) = \tan (A/2) \tan (AMC/2) \tan (B/2) \tan (CMB/2)</math>.<br />
</center><br />
<br />
But this follows from the fact that the angles <math>AMC</math> and <math>CBM</math> are supplementary.<br />
<br />
==Solution 2==<br />
By similar triangles and the fact that both centers lie on the angle bisector of <math>\angle{C}</math>, we have <math>\frac{r}{q} = \frac{s-c}{s} = \frac{a + b - c}{a + b + c}</math>, where <math>s</math> is the semi-perimeter of <math>ABC</math>. Let <math>ABC</math> have sides <math>a, b, c</math>, and let <math>AM = c_1, MB = c_2, MC = d</math>. After simple computations, we see that the condition, whose equivalent form is<br />
<cmath>\frac{b + d - c_1}{b + d + c_1} \cdot \frac{a + d - c_2}{a + d + c_2} = \frac{a + b - c}{a + b + c},</cmath><br />
is also equivalent to Stewart's Inequality<br />
<cmath>d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.</cmath><br />
{{alternate solutions}}<br />
<br />
{{IMO box|year=1970|before=First question|num-a=2}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_3&diff=703272005 USAMO Problems/Problem 32015-05-20T01:30:19Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
(''Zuming Feng'') Let <math>ABC</math> be an acute-angled triangle, and let <math>P</math> and <math>Q</math> be two points on side <math>BC</math>. Construct point <math>C_1 </math> in such a way that convex quadrilateral <math>APBC_1</math> is cyclic, <math>QC_1 \parallel CA</math>, and <math>C_1</math> and <math>Q</math> lie on opposite sides of line <math>AB</math>. Construct point <math>B_1</math> in such a way that convex quadrilateral <math>APCB_1</math> is cyclic, <math>QB_1 \parallel BA </math>, and <math>B_1 </math> and <math>Q </math> lie on opposite sides of line <math>AC</math>. Prove that points <math>B_1, C_1,P</math>, and <math>Q</math> lie on a circle.<br />
<br />
== Solution ==<br />
Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>. It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>.<br />
<br />
<asy><br />
size(300);<br />
defaultpen(1);<br />
<br />
pair A=(2,5), B=(-1,0), C=(5,0);<br />
pair C1=(.5,5.7);<br />
path O1=circumcircle(A,B,C1);<br />
pair P=IntersectionPoint(O1,B--C,1);<br />
path O2=circumcircle(A,P,C);<br />
pair B1=IntersectionPoint(O2,C1--5A-4C1,0);<br />
path O=circumcircle(B1,C1,P);<br />
pair Q=IntersectionPoint(O,B--C,1);<br />
<br />
draw(C1--P--A--B--C--A);<br />
draw(P--B1--C1--Q--B1);<br />
draw(O1,dashed+linewidth(.7));<br />
draw(O2,dashed+linewidth(.7));<br />
draw(O,dotted);<br />
<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$P$",P,S);<br />
label("$Q'$",Q,S);<br />
label("$C_1$",C1,N);<br />
label("$B_1'$",B1,E);<br />
</asy><br />
<br />
Since points <math>C_1, P, Q', B_1'</math> are concyclic and points <math>C_1, A,B_1'</math> are collinear, it follows that<br />
<cmath> \angle C_1 Q' P \equiv \angle C_1 B_1' P \equiv \angle A B_1' P . </cmath><br />
But since points <math>A, B_1', P, C</math> are concyclic,<br />
<cmath> \angle AB_1'P \equiv \angle ACP . </cmath><br />
It follows that lines <math>AC</math> and <math>C_1 Q'</math> are parallel. If we exchange <math>C</math> with <math>B</math> and <math>C_1</math> with <math>B_1'</math> in this argument, we see that lines <math>AB</math> and <math>B_1' Q'</math> are likewise parallel.<br />
<br />
It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>. Hence <math>Q' = Q</math>. Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>. Hence <math>B_1' = B_1</math>, as desired. <math>\blacksquare</math><br />
<br />
==Solution 2==<br />
''Lemma''. <math>B_1, A, C_1</math> are collinear.<br />
<br />
Suppose they are not collinear. Let line <math>B_1 A</math> intersect circle <math>ABP</math> (i.e. the circumcircle of <math>ABP</math>) again at <math>C_2</math> distinct from <math>C_1</math>. Because <math>\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ</math>, we have that <math>B_1 C_2 PQ</math> is cyclic. Hence <math>\angle C_2 QP = \angle C_2 B_1 P = \angle P B_1 A = \angle C</math>, so <math>C_2 Q // AC</math>. Then <math>C_2</math> must be the other intersection of the parallel to <math>AC</math> through <math>Q</math> with circle <math>ABP</math>. Then <math>C_2</math> is on segment <math>C_1 Q</math>, so <math>C_2</math> is contained in triangle <math>ABQ</math>. However, line <math>AB_1</math> intersects this triangle only at point <math>A</math> because <math>B_1</math> lies on arc <math>AC</math> not containing <math>P</math> of circle <math>APC</math>, a contradiction. Hence, <math>B_1, A, C_1</math> are collinear, as desired.<br />
<br />
As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired.<br />
<br />
{{alternate solutions}}<br />
<br />
== See also ==<br />
* <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url><br />
<br />
{{USAMO newbox|year=2005|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_3&diff=703262005 USAMO Problems/Problem 32015-05-20T01:29:29Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
(''Zuming Feng'') Let <math>ABC</math> be an acute-angled triangle, and let <math>P</math> and <math>Q</math> be two points on side <math>BC</math>. Construct point <math>C_1 </math> in such a way that convex quadrilateral <math>APBC_1</math> is cyclic, <math>QC_1 \parallel CA</math>, and <math>C_1</math> and <math>Q</math> lie on opposite sides of line <math>AB</math>. Construct point <math>B_1</math> in such a way that convex quadrilateral <math>APCB_1</math> is cyclic, <math>QB_1 \parallel BA </math>, and <math>B_1 </math> and <math>Q </math> lie on opposite sides of line <math>AC</math>. Prove that points <math>B_1, C_1,P</math>, and <math>Q</math> lie on a circle.<br />
<br />
== Solution ==<br />
Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>. It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>.<br />
<br />
<asy><br />
size(300);<br />
defaultpen(1);<br />
<br />
pair A=(2,5), B=(-1,0), C=(5,0);<br />
pair C1=(.5,5.7);<br />
path O1=circumcircle(A,B,C1);<br />
pair P=IntersectionPoint(O1,B--C,1);<br />
path O2=circumcircle(A,P,C);<br />
pair B1=IntersectionPoint(O2,C1--5A-4C1,0);<br />
path O=circumcircle(B1,C1,P);<br />
pair Q=IntersectionPoint(O,B--C,1);<br />
<br />
draw(C1--P--A--B--C--A);<br />
draw(P--B1--C1--Q--B1);<br />
draw(O1,dashed+linewidth(.7));<br />
draw(O2,dashed+linewidth(.7));<br />
draw(O,dotted);<br />
<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$P$",P,S);<br />
label("$Q'$",Q,S);<br />
label("$C_1$",C1,N);<br />
label("$B_1'$",B1,E);<br />
</asy><br />
<br />
Since points <math>C_1, P, Q', B_1'</math> are concyclic and points <math>C_1, A,B_1'</math> are collinear, it follows that<br />
<cmath> \angle C_1 Q' P \equiv \angle C_1 B_1' P \equiv \angle A B_1' P . </cmath><br />
But since points <math>A, B_1', P, C</math> are concyclic,<br />
<cmath> \angle AB_1'P \equiv \angle ACP . </cmath><br />
It follows that lines <math>AC</math> and <math>C_1 Q'</math> are parallel. If we exchange <math>C</math> with <math>B</math> and <math>C_1</math> with <math>B_1'</math> in this argument, we see that lines <math>AB</math> and <math>B_1' Q'</math> are likewise parallel.<br />
<br />
It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>. Hence <math>Q' = Q</math>. Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>. Hence <math>B_1' = B_1</math>, as desired. <math>\blacksquare</math><br />
<br />
==Solution 2==<br />
''Lemma''. <math>B_1, A, C_1</math> are collinear.<br />
<br />
Suppose they are not collinear. Let line <math>B_1 A</math> intersect circle <math>ABP</math> (i.e. the circumcircle of <math>ABP</math>) at <math>C_2</math> distinct from <math>C_1</math>. Because <math>\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ</math>, we have that <math>B_1 C_2 PQ</math> is cyclic. Hence <math>\angle C_2 QP = \angle C_2 B_1 P = \angle P B_1 A = \angle C</math>, so <math>C_2 Q // AC</math>. Then <math>C_2</math> must be the other intersection of the parallel to <math>AC</math> through <math>Q</math> with circle <math>ABP</math>. Then <math>C_2</math> is on segment <math>C_1 Q</math>, so <math>C_2</math> is contained in triangle <math>ABQ</math>. However, line <math>AB_1</math> intersects this triangle only at point <math>A</math> because <math>B_1</math> lies on arc <math>AC</math> not containing <math>P</math> of circle <math>APC</math>, a contradiction. Hence, <math>B_1, A, C_1</math> are collinear, as desired.<br />
<br />
As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired.<br />
<br />
{{alternate solutions}}<br />
<br />
== See also ==<br />
* <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url><br />
<br />
{{USAMO newbox|year=2005|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_3&diff=703252005 USAMO Problems/Problem 32015-05-20T01:23:45Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
(''Zuming Feng'') Let <math>ABC</math> be an acute-angled triangle, and let <math>P</math> and <math>Q</math> be two points on side <math>BC</math>. Construct point <math>C_1 </math> in such a way that convex quadrilateral <math>APBC_1</math> is cyclic, <math>QC_1 \parallel CA</math>, and <math>C_1</math> and <math>Q</math> lie on opposite sides of line <math>AB</math>. Construct point <math>B_1</math> in such a way that convex quadrilateral <math>APCB_1</math> is cyclic, <math>QB_1 \parallel BA </math>, and <math>B_1 </math> and <math>Q </math> lie on opposite sides of line <math>AC</math>. Prove that points <math>B_1, C_1,P</math>, and <math>Q</math> lie on a circle.<br />
<br />
== Solution ==<br />
Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>. It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>.<br />
<br />
<asy><br />
size(300);<br />
defaultpen(1);<br />
<br />
pair A=(2,5), B=(-1,0), C=(5,0);<br />
pair C1=(.5,5.7);<br />
path O1=circumcircle(A,B,C1);<br />
pair P=IntersectionPoint(O1,B--C,1);<br />
path O2=circumcircle(A,P,C);<br />
pair B1=IntersectionPoint(O2,C1--5A-4C1,0);<br />
path O=circumcircle(B1,C1,P);<br />
pair Q=IntersectionPoint(O,B--C,1);<br />
<br />
draw(C1--P--A--B--C--A);<br />
draw(P--B1--C1--Q--B1);<br />
draw(O1,dashed+linewidth(.7));<br />
draw(O2,dashed+linewidth(.7));<br />
draw(O,dotted);<br />
<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$P$",P,S);<br />
label("$Q'$",Q,S);<br />
label("$C_1$",C1,N);<br />
label("$B_1'$",B1,E);<br />
</asy><br />
<br />
Since points <math>C_1, P, Q', B_1'</math> are concyclic and points <math>C_1, A,B_1'</math> are collinear, it follows that<br />
<cmath> \angle C_1 Q' P \equiv \angle C_1 B_1' P \equiv \angle A B_1' P . </cmath><br />
But since points <math>A, B_1', P, C</math> are concyclic,<br />
<cmath> \angle AB_1'P \equiv \angle ACP . </cmath><br />
It follows that lines <math>AC</math> and <math>C_1 Q'</math> are parallel. If we exchange <math>C</math> with <math>B</math> and <math>C_1</math> with <math>B_1'</math> in this argument, we see that lines <math>AB</math> and <math>B_1' Q'</math> are likewise parallel.<br />
<br />
It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>. Hence <math>Q' = Q</math>. Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>. Hence <math>B_1' = B_1</math>, as desired. <math>\blacksquare</math><br />
<br />
==Solution 2==<br />
''Lemma''. <math>B_1, A, C_1</math> are collinear.<br />
<br />
Suppose they are not collinear. Let line <math>B_1 A</math> intersect circle <math>ABP</math> (i.e. the circumcircle of <math>ABP</math>) at <math>C_2</math> distinct from <math>C_1</math>. Because <math>\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ</math>, we have that <math>B_1 C_1 PQ</math> is cyclic. Hence <math>\angle C_1 QP = \angle C_1 B_1 P = \angle P B_1 A = \angle C</math>, so <math>C_2 Q // AC</math>. Then <math>C_1 = C_2</math>, contradiction. Hence, <math>B_1, A, C_1</math> are collinear.<br />
<br />
As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired.<br />
<br />
{{alternate solutions}}<br />
<br />
== See also ==<br />
* <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url><br />
<br />
{{USAMO newbox|year=2005|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1982_USAMO_Problems/Problem_3&diff=703091982 USAMO Problems/Problem 32015-05-19T03:29:00Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
If a point <math>A_1</math> is in the interior of an equilateral triangle <math>ABC</math> and point <math>A_2</math> is in the interior of <math>\triangle{A_1BC}</math>, prove that<br />
<br />
<math>I.Q. (A_1BC) > I.Q.(A_2BC)</math>,<br />
<br />
where the ''isoperimetric quotient'' of a figure <math>F</math> is defined by<br />
<br />
<math>I.Q.(F) = \frac{\text{Area (F)}}{\text{[Perimeter (F)]}^2}</math><br />
<br />
== Solution ==<br />
First, an arbitrary triangle <math>ABC</math> has isoperimetric quotient (using the notation <math>[ABC]</math> for area and <math>s = \frac{a + b + c}{2}</math>):<br />
<cmath>\frac{[ABC]}{4s^2} = \frac{[ABC]^3}{4s^2 [ABC]^2} = \frac{r^3 s^3}{4s^2 \cdot s(s-a)(s-b)(s-c)} = \frac{r^3}{4(s-a)(s-b)(s-c)}</cmath><br />
<cmath>= \frac{1}{4} \cdot \frac{r}{s-a} \cdot \frac{r}{s-b} \cdot \frac{r}{s-c} = \frac{1}{4} \tan A/2 \tan B/2 \tan C/2.</cmath><br />
<br />
Lemma. <math>\tan x \tan (A - x)</math> is increasing on <math>0 < x < \frac{A}{2}</math>, where <math>0 < A < 90^\circ</math>.<br />
<br />
Proof. <cmath>\tan x \tan (A - x) = \tan x \cdot \frac{\tan A - \tan x}{1 + \tan A \tan x} = 1 - \frac{1}{\cos^2 x (1 + \tan A \tan x)}</cmath><br />
<cmath>= 1 - \frac{2}{1 + \cos 2x + \tan A \sin 2x} = 1 - \frac{2}{1 + \sec A \cos (A - 2x)}</cmath> is increasing on the desired interval, because <math>\cos (A - 2x)</math> is increasing on <math>0 < x < \frac{A}{2}.</math><br />
<br />
Let <math>x_1, y_1, z_1</math> and <math>x_2, y_2, z_2</math> be half of the angles of triangles <math>A_1 BC</math> and <math>A_2 BC</math> in that order, respectively. Then it is immediate that <math>30^\circ > y_1 > y_2</math>, <math>30^\circ > z_1 > z_2</math>, and <math>x_1 + y_1 + z_1 = x_2 + y_2 + z_2 = 90^\circ</math>. Hence, by Lemma it follows that<br />
<cmath>\tan x_1 \tan y_1 \tan z_1 = \tan (90^\circ - y_1 - z_1) \tan y_1 \tan z_1 > \tan (90^\circ - y_1 - z_2) \tan y_1 \tan z_2</cmath><br />
<cmath> > \tan (90^\circ - y_2 - z_2) \tan y_2 \tan z_2 = \tan x_2 \tan y_2 \tan z_2.</cmath><br />
Multiplying this inequality by <math>\frac{1}{4}</math> gives that <math>I.Q[A_1 BC] > I.Q[A_2 BC]</math>, as desired.<br />
<br />
== See Also ==<br />
{{USAMO box|year=1982|num-b=2|num-a=4}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1982_USAMO_Problems/Problem_3&diff=703081982 USAMO Problems/Problem 32015-05-19T03:28:26Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
If a point <math>A_1</math> is in the interior of an equilateral triangle <math>ABC</math> and point <math>A_2</math> is in the interior of <math>\triangle{A_1BC}</math>, prove that<br />
<br />
<math>I.Q. (A_1BC) > I.Q.(A_2BC)</math>,<br />
<br />
where the ''isoperimetric quotient'' of a figure <math>F</math> is defined by<br />
<br />
<math>I.Q.(F) = \frac{\text{Area (F)}}{\text{[Perimeter (F)]}^2}</math><br />
<br />
== Solution ==<br />
First, an arbitrary triangle <math>ABC</math> has isoperimetric quotient (using the notation <math>[ABC]</math> for area and <math>s = \frac{a + b + c}{2}</math>):<br />
<cmath>\frac{[ABC]}{4s^2} = \frac{[ABC]^3}{4s^2 [ABC]^2} = \frac{r^3 s^3}{4s^2 \cdot s(s-a)(s-b)(s-c)} = \frac{r^3}{4(s-a)(s-b)(s-c)}</cmath><br />
<cmath>= \frac{1}{4} \cdot \frac{r}{s-a} \cdot \frac{r}{s-b} \cdot \frac{r}{s-c} = \frac{1}{4} \tan A/2 \tan B/2 \tan C/2.</cmath><br />
<br />
Lemma. <math>\tan x \tan (A - x)</math> is increasing on <math>0 < x < \frac{A}{2}</math>, where <math>0 < A < 90^\circ</math>.<br />
<br />
Proof. <cmath>\tan x \tan (A - x) = \tan x \cdot \frac{\tan A - \tan x}{1 + \tan A \tan x} = 1 - \frac{1}{\cos^2 x (1 + \tan A \tan x)}</cmath><br />
<cmath>= 1 - \frac{2}{1 + \cos 2x + \tan A \sin 2x} = 1 - \frac{2}{1 + \sec A \cos (A - 2x)</cmath> is increasing on the desired interval, because <math>\cos (A - 2x)</math> is increasing on <math>0 < x < \frac{A}{2}.</math><br />
<br />
Let <math>x_1, y_1, z_1</math> and <math>x_2, y_2, z_2</math> be half of the angles of triangles <math>A_1 BC</math> and <math>A_2 BC</math> in that order, respectively. Then it is immediate that <math>30^\circ > y_1 > y_2</math>, <math>30^\circ > z_1 > z_2</math>, and <math>x_1 + y_1 + z_1 = x_2 + y_2 + z_2 = 90^\circ</math>. Hence, by Lemma it follows that<br />
<cmath>\tan x_1 \tan y_1 \tan z_1 = \tan (90^\circ - y_1 - z_1) \tan y_1 \tan z_1 > \tan (90^\circ - y_1 - z_2) \tan y_1 \tan z_2</cmath><br />
<cmath> > \tan (90^\circ - y_2 - z_2) \tan y_2 \tan z_2 = \tan x_2 \tan y_2 \tan z_2.</cmath><br />
Multiplying this inequality by <math>\frac{1}{4}</math> gives that <math>I.Q[A_1 BC] > I.Q[A_2 BC]</math>, as desired.<br />
<br />
== See Also ==<br />
{{USAMO box|year=1982|num-b=2|num-a=4}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2006_USAMO_Problems/Problem_6&diff=703052006 USAMO Problems/Problem 62015-05-19T02:46:41Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
(''Zuming Feng, Zhonghao Ye'') Let <math>ABCD</math> be a quadrilateral, and let <math>E</math> and <math>F</math> be points on sides <math>AD</math> and <math>BC</math>, respectively, such that <math>AE/ED = BF/FC</math>. Ray <math>FE</math> meets rays <math>BA</math> and <math>CD</math> at <math>S</math> and <math>T</math> respectively. Prove that the circumcircles of triangles <math>SAE</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through a common point.<br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
Let the intersection of the circumcircles of <math>SAE</math> and <math>SBF</math> be <math>X</math>, and let the intersection of the circumcircles of <math>TCF</math> and <math>TDE</math> be <math>Y</math>.<br />
<br />
<math>BXF=BSF=AXE</math> because <math>BSF</math> tends both arcs <math>AE</math> and <math>BF</math>.<br />
<math>BFX=XSB=XEA</math> because <math>XSB</math> tends both arcs <math>XA</math> and <math>XB</math>.<br />
Thus, <math>XAE\sim XBF</math> by AA similarity, and <math>X</math> is the center of spiral similarity for <math>A,E,B,</math> and <math>F</math>.<br />
<math>FYC=FTC=EYD</math> because <math>FTC</math> tends both arcs <math>ED</math> and <math>FC</math>.<br />
<math>FCY=FTY=EDY</math> because <math>FTY</math> tends both arcs <math>YF</math> and <math>YE</math>.<br />
Thus, <math>YED\sim YFC</math> by AA similarity, and <math>Y</math> is the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>.<br />
<br />
From the similarity, we have that <math>XE/XF=AE/BF</math>. But we are given <math>ED/AE=CF/BF</math>, so multiplying the 2 equations together gets us <math>ED/FC=XE/XF</math>. <math>DEX,CFX</math> are the supplements of <math>AEX, BFX</math>, which are congruent, so <math>DEX=CFX</math>, and so <math>XED\sim XFC</math> by SAS similarity, and so <math>X</math> is also the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>. Thus, <math>X</math> and <math>Y</math> are the same point, which all the circumcircles pass through, and so the statement is true.<br />
<br />
=== Solution 2 ===<br />
We will give a solution using complex coordinates. The first step is the following lemma.<br />
<br />
'''Lemma.''' Suppose <math>s</math> and <math>t</math> are real numbers and <math>x</math>, <math>y</math> and <math>z</math> are complex. The circle in the complex plane passing through <math>x</math>, <math>x + ty</math> and <math>x + (s + t)z</math> also passes through the point <math>x + syz/(y - z)</math>, independent of <math>t</math>.<br />
<br />
''Proof.'' Four points <math>z_1</math>, <math>z_2</math>, <math>z_3</math> and <math>z_4</math> in the complex plane lie on a circle if and only if the cross-ratio<br />
<cmath>cr(z_1, z_2, z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}</cmath><br />
is real. Since we compute<br />
<cmath>cr(x, x + ty, x + (s + t)z, x + syz/(y - z)) = \frac{s + t}{s}</cmath><br />
the given points are on a circle. <math>\blacksquare</math><br />
<br />
Lay down complex coordinates with <math>S = 0</math> and <math>E</math> and <math>F</math> on the positive real axis. Then there are real <math>r_1</math>, <math>r_2</math> and <math>R</math> with <math>B = r_1A</math>, <math>F = r_2E</math> and <math>D = E + R(A - E)</math> and hence <math>AE/ED = BF/FC</math> gives<br />
<cmath>C = F + R(B - F) = r_2(1 - R)E + r_1RA.</cmath><br />
The line <math>CD</math> consists of all points of the form <math>sC + (1 - s)D</math> for real <math>s</math>. Since <math>T</math> lies on this line and has zero imaginary part, we see from <math>\text{Im}(sC + (1 - s)D) = (sr_1R + (1 - s)R)\text{Im}(A)</math> that it corresponds to <math>s = -1/(r_1 - 1)</math>. Thus<br />
<cmath>T = \frac{r_1D - C}{r_1 - 1} = \frac{(r_2 - r_1)(R - 1)E}{r_1 - 1}.</cmath><br />
Apply the lemma with <math>x = E</math>, <math>y = A - E</math>, <math>z = (r_2 - r_1)E/(r_1 - 1)</math>, and <math>s = (r_2 - 1)(r_1 - r_2)</math>. Setting <math>t = 1</math> gives<br />
<cmath>(x, x + y, x + (s + 1)z) = (E, A, S = 0)</cmath><br />
and setting <math>t = R</math> gives<br />
<cmath>(x, x + Ry, x + (s + R)z) = (E, D, T).</cmath><br />
Therefore the circumcircles to <math>SAE</math> and <math>TDE</math> meet at<br />
<cmath>x + \frac{syz}{y - z} = \frac{AE(r_1 - r_2)}{(1 - r_1)E - (1 - r_2)A} = \frac{AF - BE}{A + F - B - E}.</cmath><br />
This last expression is invariant under simultaneously interchanging <math>A</math> and <math>B</math> and interchanging <math>E</math> and <math>F</math>. Therefore it is also the intersection of the circumcircles of <math>SBF</math> and <math>TCF</math>.<br />
<br />
==Solution 3==<br />
Let <math>M</math> be the Miquel point of <math>ABCD</math>; then <math>M</math> is the center of the spiral similarity that takes <math>AD</math> to <math>BC</math>. Because <math>\frac{AE}{ED} = \frac{BF}{FC}</math>, the same spiral similarity also takes <math>E</math> to <math>F</math>, so <math>M</math> is the center of the spiral similarity that maps <math>AE</math> to <math>BF</math> and <math>ED</math> to <math>FC</math>. Then it is obvious that the circumcircles of <math>SAE</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through <math>M</math>.<br />
<br />
{{alternate solutions}}<br />
<br />
== See also ==<br />
* <url>viewtopic.php?t=84559 Discussion on AoPS/MathLinks</url><br />
<br />
{{USAMO newbox|year=2006|num-b=5|after=Last Question}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1999_USAMO_Problems/Problem_6&diff=702881999 USAMO Problems/Problem 62015-05-18T02:17:05Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>ABCD</math> be an isosceles trapezoid with <math>AB \parallel CD</math>. The inscribed circle <math>\omega</math> of triangle <math>BCD</math> meets <math>CD</math> at <math>E</math>. Let <math>F</math> be a point on the (internal) angle bisector of <math>\angle DAC</math> such that <math>EF \perp CD</math>. Let the circumscribed circle of triangle <math>ACF</math> meet line <math>CD</math> at <math>C</math> and <math>G</math>. Prove that the triangle <math>AFG</math> is isosceles.<br />
<br />
== Solution ==<br />
Quadrilateral <math>ABCD</math> is cyclic since it is an isosceles trapezoid. <math>AD=BC</math>. Triangle <math>ADC</math> and triangle <math>BCD</math> are reflections of each other with respect to diameter which is perpendicular to <math>AB</math>. Let the incircle of triangle <math>ADC</math> touch <math>DC</math> at <math>K</math>. The reflection implies that <math>DK=DE</math>, which then implies that the excircle of triangle <math>ADC</math> is tangent to <math>DC</math> at <math>E</math>. Since <math>EF</math> is perpendicular to <math>DC</math> which is tangent to the excircle, this implies that <math>EF</math> passes through center of excircle of triangle <math>ADC</math>.<br />
<br />
We know that the center of the excircle lies on the angular bisector of <math>DAC</math> and the perpendicular line from <math>DC</math> to <math>E</math>. This implies that <math>F</math> is the center of the excircle. <br />
<br />
Now <math>\angle GFA=\angle GCA=\angle DCA</math>. <br />
<math>\angle ACF=90+\frac{\angle DCA}{2}</math>.<br />
This means that <math>\angle AGF=90-\frac{\angle ACD}{2}</math>. (due to cyclic quadilateral <math>ACFG</math> as given).<br />
Now <math>\angle FAG - (\angle AFG + \angle FGA)=90-\frac{\angle ACD}{2}=\angle AGF</math>.<br />
<br />
Therefore <math>\angle FAG=\angle AGF</math>. <br />
QED.<br />
<br />
== See Also ==<br />
{{USAMO newbox|year=1999|num-b=5|after=Last Question}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1981_USAMO_Problems/Problem_1&diff=702871981 USAMO Problems/Problem 12015-05-17T22:16:26Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Prove that if <math>n</math> is not a multiple of <math>3</math>, then the angle <math>\frac{\pi}{n}</math> can be trisected with ruler and compasses. <br />
<br />
== Solution ==<br />
Let <math>n=3k+1</math>. Multiply throughout by <math>\pi/3n</math>. We get <br />
<br />
<math>\frac{\pi}{3} = \frac{\pi \times k}{n} + \frac{\pi}{3n}</math><br />
<br />
Re-arranging, we get<br />
<br />
<math>\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{\pi}{3n}</math><br />
<br />
A way to interpret it is that if we know the value <math>k</math>, then the reminder angle of subtracting <math>k</math> times the given angle from <math>\frac{\pi}{3}</math> gives us <math>\frac{\pi}{3n}</math>, the desired trisected angle.<br />
<br />
This can be extended to the case when <math>n=3k+2</math> where now, the equation becomes<br />
<math>\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{2\pi}{3n}</math><br />
<br />
Hence in this case, we will have to subtract <math>k</math> times the original angle from <math>\frac{\pi}{3}</math> to get twice the the trisected angle. We can bisect it after that to get the trisected angle.<br />
<br />
==Generalization==<br />
If regular polygons of <math>m</math> sides and <math>n</math> sides can be constructed, where <math>m</math> and <math>n</math> are relatively prime integers greater than or equal to three, then regular polygons of <math>mn</math> sides can be constructed. Indeed, such a polygon can be constructed by first constructing an <math>m</math>-gon, and then creating <math>m</math> distinct <math>n</math>-gons with at least one vertex being a vertex of the <math>m</math>-gon.<br />
<br />
== See Also ==<br />
{{USAMO box|year=1981|before=First Question|num-a=2}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2010_IMO_Problems/Problem_2&diff=702852010 IMO Problems/Problem 22015-05-17T17:01:57Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Given a triangle <math>ABC</math>, with <math>I</math> as its incenter and <math>\Gamma</math> as its circumcircle, <math>AI</math> intersects <math>\Gamma</math> again at <math>D</math>. Let <math>E</math> be a point on arc <math>BDC</math>, and <math>F</math> a point on the segment <math>BC</math>, such that <math>\angle BAF=\angle CAE< \frac12\angle BAC</math>. If <math>G</math> is the midpoint of <math>IF</math>, prove that the intersection of lines <math>EI</math> and <math>DG</math> lies on <math>\Gamma</math>.<br />
<br />
''Authors: Tai Wai Ming and Wang Chongli, Hong Kong''<br />
<br />
== Solution ==<br />
Note that it suffices to prove alternatively that if <math>EI</math> meets the circle again at <math>J</math> and <math>JD</math> meets <math>IF</math> at <math>G</math>, then <math>G</math> is the midpoint of <math>IF</math>.<br />
<br />
Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>.<br />
<br />
Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well.<br />
<br />
Key Lemma. Triangles <math>DKI</math> and <math>DIJ</math> are similar.<br />
Proof. Because triangles <math>DKB</math> and <math>DBJ</math> are similar by AA Similarity (for <math>\angle{KBD}</math> and <math>\angle{BJD}</math> both intercept equally sized arcs), we have <math>BD^2 = BK \cdot BJ</math>. But we know that triangle <math>DBI</math> is isosceles (hint: prove <math>\angle{BID} = \angle{IBD}</math>), and so <math>BI^2 = BK \cdot BJ</math>. Hence, by SAS Similarity, triangles <math>DKI</math> and <math>DIJ</math> are similar, as desired.<br />
<br />
Observation 3. As a result, we have <math>\angle{KID} = \angle{IJD} = \angle{DAE} = \angle{FAD}</math>.<br />
<br />
Observation 4. <math>IK // AF</math>.<br />
<br />
Observation 5. If <math>AF</math> and <math>JD</math> intersect at <math>L</math>, then <math>AJLI</math> is cyclic.<br />
<br />
Observation 6. Because <math>\angle{ALI} = \angle{AJE} = \angle{AJC} + \angle{CJE} = \angle{B} + \angle{AEC} = \angle{B} + \angle{BAF} = \angle{AFC}</math>, we have <math>LI // FK</math>.<br />
<br />
Observation 7. <math>LIKF</math> is a parallelogram, so its diagonals bisect each other, so <math>G</math> is the midpoint of <math>FI</math>, as desired.<br />
<br />
== See Also ==<br />
{{IMO box|year=2010|num-b=1|num-a=3}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2010_IMO_Problems/Problem_2&diff=702842010 IMO Problems/Problem 22015-05-17T16:59:48Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Given a triangle <math>ABC</math>, with <math>I</math> as its incenter and <math>\Gamma</math> as its circumcircle, <math>AI</math> intersects <math>\Gamma</math> again at <math>D</math>. Let <math>E</math> be a point on arc <math>BDC</math>, and <math>F</math> a point on the segment <math>BC</math>, such that <math>\angle BAF=\angle CAE< \frac12\angle BAC</math>. If <math>G</math> is the midpoint of <math>IF</math>, prove that the intersection of lines <math>EI</math> and <math>DG</math> lies on <math>\Gamma</math>.<br />
<br />
''Authors: Tai Wai Ming and Wang Chongli, Hong Kong''<br />
<br />
== Solution ==<br />
Note that it suffices to prove alternatively that if <math>EI</math> meets the circle again at <math>J</math> and <math>JD</math> meets <math>IF</math> at <math>G</math>, then <math>G</math> is the midpoint of <math>IF</math>.<br />
<br />
Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>.<br />
Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well.<br />
<br />
Key Lemma. Triangles <math>DKI</math> and <math>DIJ</math> are similar.<br />
Proof. Because triangles <math>DKB</math> and <math>DBJ</math> are similar by AA Similarity (for <math>\angle{KBD}</math> and <math>\angle{BJD}</math> both intercept equally sized arcs), we have <math>BD^2 = BK \cdot BJ</math>. But we know that triangle <math>DBI</math> is isosceles (hint: prove <math>\angle{BID} = \angle{IBD}</math>), and so <math>BI^2 = BK \cdot BJ</math>. Hence, by SAS Similarity, triangles <math>DKI</math> and <math>DIJ</math> are similar, as desired.<br />
<br />
Observation 3. As a result, we have <math>\angle{KID} = \angle{IJD} = \angle{DAE} = \angle{FAD}</math>.<br />
<br />
Observation 4. <math>IK // AF</math>.<br />
<br />
Observation 5. If <math>AF</math> and <math>JD</math> intersect at <math>L</math>, then <math>AJLI</math> is cyclic.<br />
<br />
Observation 6. Because <math>\angle{ALI} = \angle{AJE} = \angle{AJC} + \angle{CJE} = \angle{B} + \angle{AEC} = \angle{B} + \angle{BAF} = \angle{AFC}</math>, we have <math>LI // FK</math>.<br />
<br />
Observation 7. <math>LIKF</math> is a parallelogram, so its diagonals bisect each other, so <math>G</math> is the midpoint of <math>FI</math>, as desired.<br />
<br />
== See Also ==<br />
{{IMO box|year=2010|num-b=1|num-a=3}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2010_IMO_Problems/Problem_2&diff=702832010 IMO Problems/Problem 22015-05-17T16:59:30Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
<br />
Given a triangle <math>ABC</math>, with <math>I</math> as its incenter and <math>\Gamma</math> as its circumcircle, <math>AI</math> intersects <math>\Gamma</math> again at <math>D</math>. Let <math>E</math> be a point on arc <math>BDC</math>, and <math>F</math> a point on the segment <math>BC</math>, such that <math>\angle BAF=\angle CAE< \frac12\angle BAC</math>. If <math>G</math> is the midpoint of <math>IF</math>, prove that the intersection of lines <math>EI</math> and <math>DG</math> lies on <math>\Gamma</math>.<br />
<br />
''Authors: Tai Wai Ming and Wang Chongli, Hong Kong''<br />
<br />
== Solution ==<br />
Note that it suffices to prove alternatively that if <math>EI</math> meets the circle again at <math>J</math> and <math>JD</math> meets <math>IF</math> at <math>G</math>, then <math>G</math> is the midpoint of <math>IF</math>.<br />
<br />
Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>.<br />
Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well.<br />
<br />
Key Lemma. Triangles <math>DKI</math> and <math>DIJ</math> are similar.<br />
Proof. Because triangles <math>DKB</math> and <math>DBJ</math> are similar by AA Similarity (for <math>\angle{KBD}</math> and <math>\angle{BJD}</math> both intercept equally sized arcs), we have <math>BD^2 = BK \cdot BJ</math>. But we know that triangle <math>DBI</math> is isosceles (hint: prove <math>\angle{BID} = \angle{IBD}</math>), and so <math>BI^2 = BK \cdot BJ</math>. Hence, by SAS Similarity, triangles <math>DKI</math> and <math>DIJ</math> are similar, as desired.<br />
<br />
Observation 3. As a result, we have <math>\angle{KID} = \angle{IJD} = \angle{DAE} = \angle{FAD}</math>.<br />
<br />
Observation 4. <math>IK // AF</math>.<br />
<br />
Observation 5. If <math>AF</math> and <math>JD</math> intersect at <math>L</math>, then <math>AJLI</math> is cyclic.<br />
<br />
Observation 6. Because <math>\angle{ALI} = \angle{AJE} = \angle{AJC} + \angle{CJE} = \angle{B} + \angle{AEC} = \angle{B} + \angle{BAF} = \angle{AFC}, we have </math>LI // FK<math>.<br />
<br />
Observation 7. </math>LIKF<math> is a parallelogram, so its diagonals bisect each other, so </math>G<math> is the midpoint of </math>FI$, as desired.<br />
<br />
== See Also ==<br />
{{IMO box|year=2010|num-b=1|num-a=3}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1966_IMO_Problems/Problem_6&diff=702721966 IMO Problems/Problem 62015-05-17T04:37:32Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In the interior of sides <math>BC, CA, AB</math> of triangle <math>ABC</math>, any points <math>K, L,M</math>, respectively, are selected. Prove that the area of at least one of the triangles <math>AML, BKM, CLK</math> is less than or equal to one quarter of the area of triangle <math>ABC</math>.<br />
<br />
== Solution ==<br />
Let the lengths of sides <math>BC</math>, <math>CA</math>, and <math>AB</math> be <math>a</math>, <math>b</math>, and <math>c</math>, respectively. Let <math>BK=d</math>, <math>CL=e</math>, and <math>AM=f</math>.<br />
<br />
Now assume for the sake of contradiction that the areas of <math>\Delta AML</math>, <math>\Delta BKM</math>, and <math>\Delta CLK</math> are all at greater than one fourth of that of <math>\Delta ABC</math>. Therefore<br />
<br />
<cmath>\frac{AM\cdot AL\sin{\angle BAC}}{2}>\frac{AB\cdot AC\sin{\angle BAC}}{8}</cmath><br />
<br />
In other words, <math>AM\cdot AL>\frac{1}{4}AB\cdot AC</math>, or <math>f(b-e)>\frac{bc}{4}</math>. Similarly, <math>d(c-f)>\frac{ac}{4}</math> and <math>e(a-d)>\frac{ab}{4}</math>. Multiplying these three inequalities together yields<br />
<br />
<cmath>def(a-d)(b-e)(c-f)>\frac{a^2b^2c^2}{64}</cmath><br />
<br />
We also have that <math>d(a-d)\leq \frac{a^2}{4}</math>, <math>e(b-e)\leq \frac{b^2}{4}</math>, and <math>f(c-f)\leq \frac{c^2}{4}</math> from the [[Arithmetic Mean-Geometric Mean Inequality]]. Multiplying these three inequalities together yields<br />
<br />
<cmath>def(a-d)(b-e)(c-f)\leq\frac{a^2b^2c^2}{64}</cmath><br />
<br />
This is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.<br />
<br />
==Solution 2==<br />
Let <math>AR : AB = x, BP : BC = y, CQ : CA = z</math>. Then it is clear that the ratio of areas of <math>AQR, BPR, CPQ</math> to that of <math>ABC</math> equals <math>x(1-y), y(1-z), z(1-x)</math>, respectively. Suppose all three quantities exceed <math>\frac{1}{4}</math>. Then their product also exceeds <math>\frac{1}{64}</math>. However, it is clear by AM-GM that <math>x(1-x) \le \frac{1}{4}</math>, and so the product of all three quantities cannot exceed <math>\frac{1}{64}</math> (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to <math>\frac{1}{4} [ABC]</math>.<br />
<br />
== See Also ==<br />
{{IMO box|year=1966|num-b=5|after=Last Problem}}<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1960_IMO_Problems/Problem_7&diff=702711960 IMO Problems/Problem 72015-05-17T04:23:48Z<p>Suli: /* Solution */</p>
<hr />
<div>==Problem==<br />
An isosceles trapezoid with bases <math>a</math> and <math>c</math> and altitude <math>h</math> is given.<br />
<br />
a) On the axis of symmetry of this trapezoid, find all points <math>P</math> such that both legs of the trapezoid subtend right angles at <math>P</math>;<br />
<br />
b) Calculate the distance of <math>P</math> from either base;<br />
<br />
c) Determine under what conditions such points <math>P</math> actually exist. Discuss various cases that might arise.<br />
<br />
==Solution==<br />
(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.<br />
<br />
(b) Let <math>x</math> be the distance from <math>P</math> to one of the bases; then <math>h - x</math> must be the distance from <math>P</math> to the other base. Similar triangles give <math>\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}</math>, so <math>x^2 - hx + \frac{ac}{4} = 0</math> and so <math>x = \frac{h \pm \sqrt{h^2 - ac}}{2}.</math><br />
<br />
(c) When <math>h^2 \ge ac</math>.<br />
<br />
==See Also==<br />
<br />
{{IMO7 box|year=1960|num-b=6|after=Last Question}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
<br />
<center><br />
[https://www.artofproblemsolving.com/Wiki/images/0/01/1960_7.jpg]<br />
</center><br />
<br />
In our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0). <br />
<br />
Let our point P be on the axis of symmetry at z distance from the origin O. <br />
<br />
The coordinates of the points A,B,C,D,E,F and P are given in the figure. <br />
<br />
Now, <br />
<br />
Slope of the line PC= (z-0)/(0-c/2) = -2z/c<br />
Slope of the line PB= (z-h)/(0-a/2) = -2(z-h)/a<br />
<br />
Since the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1. <br />
<br />
i.e <br />
<br />
4z(z-h)=-ac<br />
<br />
or z^2 - zh + ac/4= O<br />
<br />
Now, solving for z, we get, z= [(h + ( h^2 - ac ) ^1/2 ]/2 and [(h - ( h^2 - ac ) ^1/2 ]/2<br />
<br />
So, z is the distance of the points from the base CD.. <br />
<br />
Also the points are possible only when , h^2 - ac >= 0.. and doesn't exist for h^2 -ac <0</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1960_IMO_Problems/Problem_7&diff=702701960 IMO Problems/Problem 72015-05-17T04:23:17Z<p>Suli: /* Solution */</p>
<hr />
<div>==Problem==<br />
An isosceles trapezoid with bases <math>a</math> and <math>c</math> and altitude <math>h</math> is given.<br />
<br />
a) On the axis of symmetry of this trapezoid, find all points <math>P</math> such that both legs of the trapezoid subtend right angles at <math>P</math>;<br />
<br />
b) Calculate the distance of <math>P</math> from either base;<br />
<br />
c) Determine under what conditions such points <math>P</math> actually exist. Discuss various cases that might arise.<br />
<br />
==Solution==<br />
(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.<br />
<br />
(b)Let <math>x</math> be the distance from <math>P</math> to one of the bases; then <math>h - x</math> must be the distance from <math>P</math> to the other base. Similar triangles give <math>\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}</math>, so <math>x^2 - hx + \frac{ac}{4} = 0</math> and so <math>x = \frac{h \pm \sqrt{h^2 - ac}}{2}.</math><br />
<br />
(c) When <math>h^2 \ge ac</math>.<br />
<br />
==See Also==<br />
<br />
{{IMO7 box|year=1960|num-b=6|after=Last Question}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
<br />
<center><br />
[https://www.artofproblemsolving.com/Wiki/images/0/01/1960_7.jpg]<br />
</center><br />
<br />
In our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0). <br />
<br />
Let our point P be on the axis of symmetry at z distance from the origin O. <br />
<br />
The coordinates of the points A,B,C,D,E,F and P are given in the figure. <br />
<br />
Now, <br />
<br />
Slope of the line PC= (z-0)/(0-c/2) = -2z/c<br />
Slope of the line PB= (z-h)/(0-a/2) = -2(z-h)/a<br />
<br />
Since the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1. <br />
<br />
i.e <br />
<br />
4z(z-h)=-ac<br />
<br />
or z^2 - zh + ac/4= O<br />
<br />
Now, solving for z, we get, z= [(h + ( h^2 - ac ) ^1/2 ]/2 and [(h - ( h^2 - ac ) ^1/2 ]/2<br />
<br />
So, z is the distance of the points from the base CD.. <br />
<br />
Also the points are possible only when , h^2 - ac >= 0.. and doesn't exist for h^2 -ac <0</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1960_IMO_Problems/Problem_7&diff=702691960 IMO Problems/Problem 72015-05-17T04:23:03Z<p>Suli: </p>
<hr />
<div>==Problem==<br />
An isosceles trapezoid with bases <math>a</math> and <math>c</math> and altitude <math>h</math> is given.<br />
<br />
a) On the axis of symmetry of this trapezoid, find all points <math>P</math> such that both legs of the trapezoid subtend right angles at <math>P</math>;<br />
<br />
b) Calculate the distance of <math>P</math> from either base;<br />
<br />
c) Determine under what conditions such points <math>P</math> actually exist. Discuss various cases that might arise.<br />
<br />
==Solution==<br />
(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.<br />
<br />
(b)Let <math>x</math> be the distance from <math>P</math> to one of the bases; then <math>h - x</math> must be the distance from <math>P</math> to the other base. Similar triangles give <math>\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}, so </math>x^2 - hx + \frac{ac}{4} = 0<math> and so </math>x = \frac{h \pm \sqrt{h^2 - ac}}{2}.<math><br />
<br />
(c) When </math>h^2 \ge ac$.<br />
<br />
==See Also==<br />
<br />
{{IMO7 box|year=1960|num-b=6|after=Last Question}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
<br />
<center><br />
[https://www.artofproblemsolving.com/Wiki/images/0/01/1960_7.jpg]<br />
</center><br />
<br />
In our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0). <br />
<br />
Let our point P be on the axis of symmetry at z distance from the origin O. <br />
<br />
The coordinates of the points A,B,C,D,E,F and P are given in the figure. <br />
<br />
Now, <br />
<br />
Slope of the line PC= (z-0)/(0-c/2) = -2z/c<br />
Slope of the line PB= (z-h)/(0-a/2) = -2(z-h)/a<br />
<br />
Since the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1. <br />
<br />
i.e <br />
<br />
4z(z-h)=-ac<br />
<br />
or z^2 - zh + ac/4= O<br />
<br />
Now, solving for z, we get, z= [(h + ( h^2 - ac ) ^1/2 ]/2 and [(h - ( h^2 - ac ) ^1/2 ]/2<br />
<br />
So, z is the distance of the points from the base CD.. <br />
<br />
Also the points are possible only when , h^2 - ac >= 0.. and doesn't exist for h^2 -ac <0</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1960_IMO_Problems/Problem_3&diff=702681960 IMO Problems/Problem 32015-05-17T04:14:17Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> an odd integer). Let <math>\alpha</math> be the acute angle subtending, from <math>A</math>, that segment which contains the midpoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse of the triangle. Prove that:<br />
<center><math><br />
\tan{\alpha}=\frac{4nh}{(n^2-1)a}.<br />
</math><br />
</center><br />
<br />
== Solution ==<br />
Using coordinates, let <math>A=(0,0)</math>, <math>B=(b,0)</math>, and <math>C=(0,c)</math>. Also, let <math>PQ</math> be the segment that contains the midpoint of the hypotenuse with <math>P</math> closer to <math>B</math>. <br />
<br />
<asy><br />
size(8cm);<br />
pair A,B,C,P,Q;<br />
A=(0,0);<br />
B=(4,0);<br />
C=(0,3);<br />
P=(2.08,1.44);<br />
Q=(1.92,1.56);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
label("A",A,SW);<br />
label("B",B,SE);<br />
label("C",C,NW);<br />
label("P",P,ENE);<br />
label("Q",Q,NNE);<br />
draw(A--B--C--cycle); <br />
draw(A--P); <br />
draw(A--Q); <br />
</asy><br />
<br />
Then, <math>P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)</math>, and <math>Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)</math>. <br />
<br />
So, <math>\text{slope}</math><math>(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}</math>, and <math>\text{slope}</math><math>(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}</math>. <br />
<br />
Thus, <math>\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}</math><br />
<math>= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}</math>. <br />
<br />
Since <math>[ABC]=\frac{1}{2}bc=\frac{1}{2}ah</math>, <math>bc=ah</math> and <math>\tan{\alpha}=\frac{4nh}{(n^2-1)a}</math> as desired. <br />
<br />
==Solution 2==<br />
Let <math>P, Q, R</math> be points on side <math>BC</math> such that segment <math>PR</math> contains midpoint <math>Q</math>, with <math>P</math> closer to <math>C</math> and (without loss of generality) <math>AC \le AB</math>. Then if <math>AD</math> is an altitude, then <math>D</math> is between <math>P</math> and <math>C</math>. Combined with the obvious fact that <math>Q</math> is the midpoint of <math>PR</math> (for <math>n</math> is odd), we have<br />
<cmath>\tan {\angle PAR} = \tan (\angle RAD - \angle PAD) = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4nh}{(n^2-1)a}.</cmath><br />
<br />
==See Also==<br />
{{IMO7 box|year=1960|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1960_IMO_Problems/Problem_3&diff=702671960 IMO Problems/Problem 32015-05-17T04:13:36Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> an odd integer). Let <math>\alpha</math> be the acute angle subtending, from <math>A</math>, that segment which contains the midpoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse of the triangle. Prove that:<br />
<center><math><br />
\tan{\alpha}=\frac{4nh}{(n^2-1)a}.<br />
</math><br />
</center><br />
<br />
== Solution ==<br />
Using coordinates, let <math>A=(0,0)</math>, <math>B=(b,0)</math>, and <math>C=(0,c)</math>. Also, let <math>PQ</math> be the segment that contains the midpoint of the hypotenuse with <math>P</math> closer to <math>B</math>. <br />
<br />
<asy><br />
size(8cm);<br />
pair A,B,C,P,Q;<br />
A=(0,0);<br />
B=(4,0);<br />
C=(0,3);<br />
P=(2.08,1.44);<br />
Q=(1.92,1.56);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
label("A",A,SW);<br />
label("B",B,SE);<br />
label("C",C,NW);<br />
label("P",P,ENE);<br />
label("Q",Q,NNE);<br />
draw(A--B--C--cycle); <br />
draw(A--P); <br />
draw(A--Q); <br />
</asy><br />
<br />
Then, <math>P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)</math>, and <math>Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)</math>. <br />
<br />
So, <math>\text{slope}</math><math>(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}</math>, and <math>\text{slope}</math><math>(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}</math>. <br />
<br />
Thus, <math>\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}</math><br />
<math>= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}</math>. <br />
<br />
Since <math>[ABC]=\frac{1}{2}bc=\frac{1}{2}ah</math>, <math>bc=ah</math> and <math>\tan{\alpha}=\frac{4nh}{(n^2-1)a}</math> as desired. <br />
<br />
==Solution 2==<br />
Let <math>P, Q, R</math> be points on side <math>BC</math> such that segment <math>PR</math> contains midpoint <math>Q</math>, with <math>P</math> closer to <math>C</math> and (without loss of generality) <math>AC \le AB</math>. Then if <math>AD</math> is an altitude, then <math>D</math> is between <math>P</math> and <math>C</math>. Combined with the obvious fact that <math>Q</math> is the midpoint of <math>PR</math> (for <math>n</math> is odd), we have<br />
<cmath>\tan {\angle PAR} = \tan {\angle RAD - \angle PAD} = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4n}{a(n^2-1)}.</cmath><br />
<br />
==See Also==<br />
{{IMO7 box|year=1960|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems/Problem_1&diff=702662001 IMO Problems/Problem 12015-05-17T03:38:11Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.<br />
<br />
== Solution ==<br />
Take <math>D</math> on the circumcircle with <math>AD \parallel BC</math>. Notice that <math>\angle CBD = \angle BCA</math>, so <math>\angle ABD \ge 30^\circ</math>. Hence <math>\angle AOD \ge 60^\circ</math>. Let <math>Z</math> be the midpoint of <math>AD</math> and <math>Y</math> the midpoint of <math>BC</math>. Then <math>AZ \ge R/2</math>, where <math>R</math> is the radius of the circumcircle. But <math>AZ = YX</math> (since <math>AZYX</math> is a rectangle).<br />
<br />
Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math>OX > YX \ge R/2</math>. But <math>XC = YC - YX < R - YX \le R/2</math>. So <math>OX > XC</math>.<br />
<br />
Hence <math>\angle COX < \angle OCX</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCX = \angle ECB</math>. But <math>\angle ECB = \angle EAB</math> and <math>\angle EAB + \angle BAC = \angle EAC = 90^\circ</math>, since <math>EC</math> is a diameter. Hence <math>\angle COX + \angle BAC < 90^\circ</math>.<br />
<br />
==Solution 2==<br />
Notice that because <math>\angle{PCO} = 90^\circ - \angle{A}</math>, it suffices to prove that <math>\angle{POC} < \angle{PCO}</math>, or equivalently <math>PC < PO.</math><br />
<br />
Suppose on the contrary that <math>PC > PO</math>. By the triangle inequality, <math>2 PC = PC + PC > PC + PO = CO = R</math>, where <math>R</math> is the circumradius of <math>ABC</math>. But the Law of Sines and basic trigonometry gives us that <math>PC = 2R \sin B \cos C</math>, so we have <math>4 \sin B \cos C > 1</math>. But we also have <math>4 \sin B \cos C \le 4 \sin B \cos (B + 30^\circ) = 2 (\sin (2B + 30^\circ) - \sin 30^\circ) \le 2 (1 - \frac{1}{2}) = 1</math> because <math>\angle{C} \ge \angle{B} + 30^\circ</math>, and so we have a contradiction. Hence <math>PC < PO</math> and so <math>\angle{PCO} + \angle{A} < 90^\circ</math>, as desired.<br />
<br />
== See also ==<br />
{{IMO box|year=2001|before=First question|num-a=2}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems/Problem_1&diff=702652001 IMO Problems/Problem 12015-05-17T03:37:39Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.<br />
<br />
== Solution ==<br />
Take <math>D</math> on the circumcircle with <math>AD \parallel BC</math>. Notice that <math>\angle CBD = \angle BCA</math>, so <math>\angle ABD \ge 30^\circ</math>. Hence <math>\angle AOD \ge 60^\circ</math>. Let <math>Z</math> be the midpoint of <math>AD</math> and <math>Y</math> the midpoint of <math>BC</math>. Then <math>AZ \ge R/2</math>, where <math>R</math> is the radius of the circumcircle. But <math>AZ = YX</math> (since <math>AZYX</math> is a rectangle).<br />
<br />
Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math>OX > YX \ge R/2</math>. But <math>XC = YC - YX < R - YX \le R/2</math>. So <math>OX > XC</math>.<br />
<br />
Hence <math>\angle COX < \angle OCX</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCX = \angle ECB</math>. But <math>\angle ECB = \angle EAB</math> and <math>\angle EAB + \angle BAC = \angle EAC = 90^\circ</math>, since <math>EC</math> is a diameter. Hence <math>\angle COX + \angle BAC < 90^\circ</math>.<br />
<br />
==Solution 2==<br />
Notice that because <math>\angle{PCO} = 90^\circ - \angle{A}</math>, it suffices to prove that <math>\angle{POC} < \angle{PCO}</math>, or equivalently <math>PC < PO.</math><br />
<br />
Suppose on the contrary that <math>PC > PO</math>. By the triangle inequality, <math>2 PC = PC + PC > PC + PO = CO = R</math>, where <math>R</math> is the circumradius of <math>ABC</math>. But the Law of Sines and basic trigonometry gives us that <math>PC = 2R \sin B \cos C</math>, so we have <math>4 \sin B \cos C \ge 1</math>. But we also have <math>4 \sin B \cos C \le 4 \sin B \cos (B + 30^\circ) = 2 (\sin (2B + 30^\circ) - \sin 30^\circ) \le 2 (1 - \frac{1}{2}) = 1</math> because <math>\angle{C} \ge \angle{B} + 30^\circ</math>, and so we have a contradiction. Hence <math>PC < PO</math> and so <math>\angle{PCO} + \angle{A} < 90^\circ</math>, as desired.<br />
<br />
== See also ==<br />
{{IMO box|year=2001|before=First question|num-a=2}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems/Problem_1&diff=702642001 IMO Problems/Problem 12015-05-17T03:26:17Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.<br />
<br />
== Solution ==<br />
Take <math>D</math> on the circumcircle with <math>AD \parallel BC</math>. Notice that <math>\angle CBD = \angle BCA</math>, so <math>\angle ABD \ge 30^\circ</math>. Hence <math>\angle AOD \ge 60^\circ</math>. Let <math>Z</math> be the midpoint of <math>AD</math> and <math>Y</math> the midpoint of <math>BC</math>. Then <math>AZ \ge R/2</math>, where <math>R</math> is the radius of the circumcircle. But <math>AZ = YX</math> (since <math>AZYX</math> is a rectangle).<br />
<br />
Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math>OX > YX \ge R/2</math>. But <math>XC = YC - YX < R - YX \le R/2</math>. So <math>OX > XC</math>.<br />
<br />
Hence <math>\angle COX < \angle OCX</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCX = \angle ECB</math>. But <math>\angle ECB = \angle EAB</math> and <math>\angle EAB + \angle BAC = \angle EAC = 90^\circ</math>, since <math>EC</math> is a diameter. Hence <math>\angle COX + \angle BAC < 90^\circ</math>.<br />
<br />
==Solution 2==<br />
Notice that because <math>\angle{PCO} = 90^\circ - \angle{A}</math>, it suffices to prove that <math>\angle{POC} < \angle{PCO}</math>, or equivalently <math>PC < PO.</math><br />
<br />
Suppose on the contrary that <math>PC > PO</math>. By the triangle inequality, <math>2 PC = PC + PC > PC + PO = CO = R</math>, where <math>R</math> is the circumradius of <math>ABC</math>. But the Law of Sines and basic trigonometry gives us that <math>PC = 2R \sin B \cos C</math>, so we have <math>2 \sin B \cos C \ge 1</math>. But we also have <math>2 \sin B \cos C < 2 \sin B \cos B = \sin 2B \le 1</math> because <math>\angle{C} > \angle{B}</math>, and so we have a contradiction. Hence <math>PC < PO</math> and so <math>\angle{PCO} + \angle{A} < 90^\circ</math>, as desired.<br />
<br />
== See also ==<br />
{{IMO box|year=2001|before=First question|num-a=2}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems/Problem_1&diff=702632001 IMO Problems/Problem 12015-05-17T03:24:46Z<p>Suli: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.<br />
<br />
== Solution ==<br />
Take <math>D</math> on the circumcircle with <math>AD \parallel BC</math>. Notice that <math>\angle CBD = \angle BCA</math>, so <math>\angle ABD \ge 30^\circ</math>. Hence <math>\angle AOD \ge 60^\circ</math>. Let <math>Z</math> be the midpoint of <math>AD</math> and <math>Y</math> the midpoint of <math>BC</math>. Then <math>AZ \ge R/2</math>, where <math>R</math> is the radius of the circumcircle. But <math>AZ = YX</math> (since <math>AZYX</math> is a rectangle).<br />
<br />
Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math>OX > YX \ge R/2</math>. But <math>XC = YC - YX < R - YX \le R/2</math>. So <math>OX > XC</math>.<br />
<br />
Hence <math>\angle COX < \angle OCX</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCX = \angle ECB</math>. But <math>\angle ECB = \angle EAB</math> and <math>\angle EAB + \angle BAC = \angle EAC = 90^\circ</math>, since <math>EC</math> is a diameter. Hence <math>\angle COX + \angle BAC < 90^\circ</math>.<br />
<br />
==Solution 2==<br />
Notice that because <math>\angle{PCO} = 90^\circ - \angle{A}</math>, it suffices to prove that <math>\angle{POC} < \angle{PCO}</math>, or equivalently <math>PC < PO.</math><br />
<br />
Suppose on the contrary that <math>PC > PO</math>. By the triangle inequality, <math>2 PC > PC + PC = PC + PO = CO = R</math>, where <math>R</math> is the circumradius of <math>ABC</math>. But the Law of Sines and basic trigonometry gives us that <math>PC = 2R \sin B \cos C</math>, so we have <math>2 \sin B \cos C \ge 1</math>. But we also have <math>2 \sin B \cos C < 2 \sin B \cos B = \sin 2B \le 1</math> because <math>\angle{C} > \angle{B}</math>, and so we have a contradiction. Hence <math>PC < PO</math> and so <math>\angle{PCO} + \angle{A} < 90^\circ</math>, as desired.<br />
<br />
== See also ==<br />
{{IMO box|year=2001|before=First question|num-a=2}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2000_USAMO_Problems/Problem_5&diff=702622000 USAMO Problems/Problem 52015-05-17T02:48:00Z<p>Suli: </p>
<hr />
<div>== Problem ==<br />
Let <math>A_1A_2A_3</math> be a [[triangle]] and let <math>\omega_1</math> be a [[circle]] in its plane passing through <math>A_1</math> and <math>A_2.</math> Suppose there exist circles <math>\omega_2, \omega_3, \dots, \omega_7</math> such that for <math>k = 2, 3, \dots, 7,</math> <math>\omega_k</math> is externally [[tangent (geometry)|tangent]] to <math>\omega_{k - 1}</math> and passes through <math>A_k</math> and <math>A_{k + 1},</math> where <math>A_{n + 3} = A_{n}</math> for all <math>n \ge 1</math>. Prove that <math>\omega_7 = \omega_1.</math><br />
<br />
== Solution ==<br />
Let the [[circumcenter]] of <math>\triangle ABC</math> be <math>O</math>, and let the center of <math>\omega_k</math> be <math>O_k</math>. <math>\omega_k</math> and <math>\omega_{k-1}</math> are externally tangent at the point <math>A_k</math>, so <math>O_k, A_k, O_{k-1}</math> are [[collinear]]. <br />
<br />
<math>O</math> is the intersection of the perpendicular bisectors of <math>\overline{A_1A_2}, \overline{A_2A_3}, \overline{A_3A_1}</math>, and each of the centers <math>O_k</math> lie on the perpendicular bisector of the side of the triangle that determines <math>\omega_k</math>. It follows from <math>OA_k = OA_{k+1}, O_kA_k = O_kA_{k+1}, OO_k = OO_k \Longrightarrow \triangle OA_kO_k \cong \triangle OA_{k+1}O_k</math> that <math>\angle OA_kO_k = \angle OA_{k+1}O_k</math>. <br />
<br />
<center><asy><br />
size(300);<br />
pathpen = linewidth(0.7); pen t = linetype("2 2");<br />
pair A = (0,0), B=3*expi(1), C=(3.5)*expi(0); /* arbitrary points */<br />
pair O=circumcenter(A,B,C), O1 = O + 5*( ((B+C)/2) - O ), O2 = IP(O -- O + 100*( ((A+C)/2) - O ), O1 -- O1 + 10*( C - O1 )); <br />
D(MP("A_3",A,SW)--MP("A_1",B,N)--MP("A_2",C,SE)--cycle); D(MP("O",O,NW)); D(MP("O_1",O1,E)); D(MP("O_2",O2)); D(O--B--O1--C--O--A--O2--C, linetype("4 4") + linewidth(0.7)); D(O1--O--O2,linetype("4 4") + linewidth(0.6)); D(CP(O1,C),t);D(CP(O2,C),t);D(circumcircle(A,B,C),t);<br />
</asy></center><br />
<br />
Since <math>O, A_k</math>, and the perpendicular bisector of <math>\overline{A_kA_{k+1}}</math> are fixed, the angle <math>OA_kO_k</math> determines the position of <math>O_k</math> (since <math>O_k</math> lies on the perpendicular bisector). Let <math>\theta_k = m\angle OA_kO_k</math>; then, <math>\theta_i = \theta_j</math> and <math>i \equiv j \pmod{3}</math> together imply that <math>O_i \equiv O_j</math>.<br />
<br />
Now <math>\theta_1 = \angle OA_1O_1 = \angle OA_2O_1 = 180 - \angle OA_2O_2 = 180 - \theta_2</math> (due to collinearility). Hence, we have the recursion <math>\theta_k = 180 - \theta_{k-1}</math>, and so <math>\theta_k = \theta_{k-2}</math>. Thus, <math>\theta_{1} = \theta_{7}</math>.<br />
<br />
<math>\theta_{1} = \theta_{7}</math> implies that <math>O_1 \equiv O_7</math>, and circles <math>\omega_1</math> and <math>\omega_7</math> are the same circle since they have the same center and go through the same two points.<br />
<br />
==Solution 2==<br />
Using the collinearity of certain points and the fact that <math>A_k A_{k+1} O_k</math> is isosceles, we quickly deduce that<br />
<cmath>\angle{A_1 A_2 O_1} = 180^\circ - \angle{A_2} - (180^\circ - \angle{A_3}) + (180^\circ - \angle{A_1}) - (180^\circ - \angle{A_2}) + (180^\circ - \angle{A_3}) - (180^\circ - \angle{A_1}) + \angle{A_7 A_8 O_7}.</cmath><br />
From ASA Congruence we deduce that <math>A_1 A_2 O_1</math> and <math>A_7 A_8 O_7</math> are congruent triangles, and so <math>O_1 A_1 = O_7 A_7</math>, that is <math>\omega_1 = \omega_7</math>.<br />
<br />
== See Also ==<br />
{{USAMO newbox|year=2000|num-b=4|num-a=6}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1987_AHSME_Problems/Problem_20&diff=699531987 AHSME Problems/Problem 202015-04-17T04:07:50Z<p>Suli: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Evaluate<br />
<math> \log_{10}(\tan 1^{\circ})+\log_{10}(\tan 2^{\circ})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). </math><br />
<br />
<math>\textbf{(A)}\ 0 \qquad<br />
\textbf{(B)}\ \frac{1}{2}\log_{10}(\frac{\sqrt{3}}{2}) \qquad<br />
\textbf{(C)}\ \frac{1}{2}\log_{10}2\qquad<br />
\textbf{(D)}\ 1\qquad<br />
\textbf{(E)}\ \text{none of these} </math><br />
<br />
==Solution==<br />
Because <math>\tan x \tan (90^\circ - x) = \tan x \cot x = 1</math>, <math>\tan 45^\circ = 1</math>, and <math>\log a + \log b = \log {ab}</math>, the answer is <math>\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.</math> <math>\boxed{\textbf{(A)}}.</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1987|num-b=19|num-a=21}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1993_USAMO_Problems/Problem_5&diff=699421993 USAMO Problems/Problem 52015-04-15T22:09:13Z<p>Suli: </p>
<hr />
<div>== Problem 5==<br />
<br />
Let <math>a_0, a_1, a_2,\cdots</math> be a sequence of positive real numbers satisfying <math>a_{i-1}a_{i+1}\le a^2_i</math><br />
for <math>i = 1, 2, 3,\cdots</math> . (Such a sequence is said to be <i>log concave</i>.) Show that for<br />
each <math>n > 1</math>,<br />
<br />
<center><math>\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}</math>.</center><br />
<br />
==Solution==<br />
Notice that because<br />
<cmath>(a_0 + a_1 + \dots + a_{n-1})(a_1 + a_2 + \dots + a_n) = (a_0 + (a_1 + \dots + a_{n-1})(-a_0 + (a_0 + a_1 + \dots + a_n))</cmath><br />
<cmath> = -a_0^2 + a_0 ((a_0 + a_1 + \dots + a_n) - (a_1 + a_2 + \dots + a_{n-1})) + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)</cmath><br />
<cmath> = a_0 a_n + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n),</cmath><br />
we may subtract <math>\dfrac{(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)}{n^2}</math> from both sides of the inequality and observe that it is sufficient to prove that<br />
<cmath>\frac{(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)}{(n^2 - 1)n^2} \ge \frac{a_0 a_n}{n^2},</cmath><br />
or<br />
<cmath>(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) a_0 a_n.</cmath><br />
<br />
Fortunately, this is an easy inequality. Indeed, from AM-GM applied on each group of terms we have<br />
<cmath>(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) \sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n},</cmath><br />
and so it suffices to prove<br />
<cmath>\sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n} \ge a_0 a_n,</cmath><br />
or, after taking both sides to the <math>(n^2 - 1)</math> power, simplifying, and taking the <math>n</math>-th root of both sides, to prove<br />
<cmath>a_1^2 a_2^2 a_3^2 \dots a_{n-1}^2 \ge a_0^{n-1} a_n^{n-1}.</cmath><br />
This easily follows from the Fact that <math>a_0 a_n \le a_i a_{n-i}</math> for <math>1 \le i \le n-1</math>. Indeed, we are given that<br />
<cmath>a_0 a_2 \le a_1^2</cmath><br />
<cmath>a_1 a_3 \le a_2^2</cmath><br />
<cmath>a_2 a_4 \le a_3^2</cmath><br />
<cmath>\dots</cmath><br />
<cmath>a_{n-3} a_{n-1} \le a_{n-2}^2</cmath><br />
<cmath>a_{n-2} a_n \le a_{n-1}^2.</cmath><br />
Multiply all inequalities together and cancel <math>a_1, a_2^2, a_3^2, \dots, a_{n-2}^2, a_{n-1}</math> to give <math>a_0 a_n \ge a_1 a_{n-1}</math>. Similarly, by multiplying all inequalities except the first and the last, we deduce that <math>a_2 a_{n-2} \le a_1 a_{n-1} \ge a_0 a_n</math>, and a simple induction argument proves the verity of the Fact for <math>1 \le i \le \frac{n}{2}</math>, and so by the Commutative Property the Fact is true for all <math>1 \le i \le n-1</math>, as desired. Now multiply each inequality of the Fact for <math>i = 1, 2, 3, \dots, n-1</math> to give the desired result.<br />
<br />
Note: The Fact can be generalized into a Lemma: <math>a_x a_w \le a_y a_z</math> whenever <math>x \le y \le z \le w</math> and <math>x + w = y + z</math>. The proof is similar to that of the Fact and is left as an exercise to the reader.<br />
<br />
--User suli, April 15, 2015.<br />
<br />
== See Also ==<br />
{{USAMO box|year=1993|num-b=4|after=Last Problem}}<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks]<br />
<br />
[[Category:Olympiad Algebra Problems]]<br />
[[Category:Olympiad Inequality Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1993_USAMO_Problems&diff=699411993 USAMO Problems2015-04-15T21:51:44Z<p>Suli: </p>
<hr />
<div>== Problem 1==<br />
<br />
For each integer <math>n\ge2</math>, determine, with proof, which of the two positive real numbers <math>a</math> and <math>b</math> satisfying<br />
<br />
<center><math>a^n = a + 1, \quad b^{2n} = b + 3a</math></center><br />
<br />
is larger.<br />
<br />
[[1993 USAMO Problems/Problem 1 | Solution]]<br />
<br />
== Problem 2==<br />
<br />
Let <math>ABCD</math> be a convex quadrilateral such that diagonals <math>AC</math> and <math>BD</math> intersect at right angles, and let <math>E</math> be their intersection. Prove that the reflections of <math>E</math> across <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DA</math> are concyclic.<br />
<br />
[[1993 USAMO Problems/Problem 2 | Solution]]<br />
<br />
== Problem 3==<br />
<br />
Consider functions <math>f : [0, 1] \rightarrow \Re</math> which satisfy<br />
<table><tr><br />
<td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td><td>(i)</td><td><math>f(x)\ge0</math> for all <math>x</math> in <math>[0, 1]</math>,</td></tr><br />
<tr><br />
<td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td><td>(ii)</td><td><math>f(1) = 1</math>,</td></tr><br />
<tr><br />
<td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td><td>(iii)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td><td><math>f(x) + f(y) \le f(x + y)</math> whenever <math>x</math>, <math>y</math>, and <math>x + y</math> are all in <math>[0, 1]</math>.</td></tr></table><br />
<br />
Find, with proof, the smallest constant <math>c</math> such that<br />
<br />
<center><math>f(x) \le cx</math></center><br />
<br />
for every function <math>f</math> satisfying (i)-(iii) and every <math>x</math> in <math>[0, 1]</math>.<br />
<br />
[[1993 USAMO Problems/Problem 3 | Solution]]<br />
<br />
== Problem 4==<br />
<br />
Let <math>a</math>, <math>b</math> be odd positive integers. Define the sequence <math>(f_n)</math> by putting <math>f_1 = a</math>,<br />
<math>f_2 = b</math>, and by letting <math>f_n</math> for <math>n\ge3</math> be the greatest odd divisor of <math>f_{n-1} + f_{n-2}</math>.<br />
Show that <math>f_n</math> is constant for <math>n</math> sufficiently large and determine the eventual<br />
value as a function of <math>a</math> and <math>b</math>.<br />
<br />
[[1993 USAMO Problems/Problem 4 | Solution]]<br />
<br />
== Problem 5==<br />
<br />
Let <math>a_0, a_1, a_2,\cdots</math> be a sequence of positive real numbers satisfying <math>a_{i-1}a_{i+1}\le a^2_i</math><br />
for <math>i = 1, 2, 3,\cdots</math> . (Such a sequence is said to be <i>log concave</i>.) Show that for<br />
each <math>n > 1</math>,<br />
<br />
<center><math>\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}</math>.</center><br />
<br />
[[1993 USAMO Problems/Problem 5 | Solution]]<br />
<br />
== See Also ==<br />
{{USAMO box|year=1993|before=[[1992 USAMO]]|after=[[1994 USAMO]]}}<br />
<br />
[[1993 USAMO Problems/Problem 5 | Solution]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1997_USAMO_Problems/Problem_5&diff=697351997 USAMO Problems/Problem 52015-03-30T03:43:14Z<p>Suli: /* Solution 3 (Isolated fudging) */</p>
<hr />
<div>== Problem ==<br />
Prove that, for all positive real numbers <math>a, b, c,</math><br />
<br />
<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>.<br />
<br />
== Solution ==<br />
[[File:USAMO97(5-solution).jpg]]<br />
<br />
== Solution 2 ==<br />
'''Outline:'''<br />
<br />
1. Because the inequality is homogenous, scale <math>a, b, c</math> by an arbitrary factor such that <math>abc = 1</math>.<br />
<br />
2. Replace all <math>abc</math> with 1. Then, multiply both sides by <math>(a^3 + b^3 + 1)(b^3 + c^3 + 1)(a^3 + c^3 + 1)</math> to clear the denominators.<br />
<br />
3. Expand each product of trinomials.<br />
<br />
4. Cancel like mad.<br />
<br />
5. You are left with <math>a^3 + a^3 + b^3 + b^3 + c^3 + c^3 \le a^6b^3 + a^6c^3 + b^6c^3 + b^6a^3 + c^6a^3 + c^6b^3</math>. Homogenize the inequality by multiplying each term of the LHS by <math>a^2b^2c^2</math>. Because <math>(6, 3, 0)</math> ''majorizes'' <math>(5, 2, 2)</math>, this inequality holds true by bunching. (Alternatively, one sees the required AM-GM is <math>\frac{a^6b^3 + a^6b^3 + a^3c^6}{3} \ge a^5b^2c^2</math>. Sum similar expressions to obtain the desired result.)<br />
<br />
==Solution 3 (Isolated fudging)==<br />
Because the inequality is homogenous (i.e. <math>(a, b, c)</math> can be replaced with <math>(ka, kb, kc)</math> without changing the inequality other than by a factor of <math>k^n</math> for some <math>n</math>), without loss of generality, let <math>abc = 1</math>.<br />
<br />
Lemma:<br />
<cmath>\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.</cmath><br />
Proof: Rearranging gives <math>(a^3 + b^3) c + c \ge a + b + c</math>, which is a simple consequence of <math>a^3 + b^3 = (a + b)(a^2 - ab + b^2)</math> and<br />
<cmath>(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.</cmath><br />
<br />
Thus, by <math>abc = 1</math>:<br />
<cmath>\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}</cmath><br />
<cmath>\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.</cmath><br />
<br />
==See Also ==<br />
{{USAMO newbox|year=1997|num-b=4|num-a=6}}<br />
<br />
[[Category:Olympiad Algebra Problems]]<br />
[[Category:Olympiad Inequality Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1997_USAMO_Problems/Problem_5&diff=697341997 USAMO Problems/Problem 52015-03-30T03:42:11Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Prove that, for all positive real numbers <math>a, b, c,</math><br />
<br />
<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>.<br />
<br />
== Solution ==<br />
[[File:USAMO97(5-solution).jpg]]<br />
<br />
== Solution 2 ==<br />
'''Outline:'''<br />
<br />
1. Because the inequality is homogenous, scale <math>a, b, c</math> by an arbitrary factor such that <math>abc = 1</math>.<br />
<br />
2. Replace all <math>abc</math> with 1. Then, multiply both sides by <math>(a^3 + b^3 + 1)(b^3 + c^3 + 1)(a^3 + c^3 + 1)</math> to clear the denominators.<br />
<br />
3. Expand each product of trinomials.<br />
<br />
4. Cancel like mad.<br />
<br />
5. You are left with <math>a^3 + a^3 + b^3 + b^3 + c^3 + c^3 \le a^6b^3 + a^6c^3 + b^6c^3 + b^6a^3 + c^6a^3 + c^6b^3</math>. Homogenize the inequality by multiplying each term of the LHS by <math>a^2b^2c^2</math>. Because <math>(6, 3, 0)</math> ''majorizes'' <math>(5, 2, 2)</math>, this inequality holds true by bunching. (Alternatively, one sees the required AM-GM is <math>\frac{a^6b^3 + a^6b^3 + a^3c^6}{3} \ge a^5b^2c^2</math>. Sum similar expressions to obtain the desired result.)<br />
<br />
==Solution 3 (Isolated fudging)==<br />
Without loss of generality, let <math>abc = 1</math>.<br />
<br />
Lemma:<br />
<cmath>\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.</cmath><br />
Proof: Rearranging gives <math>(a^3 + b^3) c + c \ge a + b + c</math>, which is a simple consequence of <math>a^3 + b^3 = (a + b)(a^2 - ab + b^2)</math> and<br />
<cmath>(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.</cmath><br />
<br />
Thus, by <math>abc = 1</math>:<br />
<cmath>\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}</cmath><br />
<cmath>\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.</cmath><br />
<br />
==See Also ==<br />
{{USAMO newbox|year=1997|num-b=4|num-a=6}}<br />
<br />
[[Category:Olympiad Algebra Problems]]<br />
[[Category:Olympiad Inequality Problems]]<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=1997_USAMO_Problems&diff=697331997 USAMO Problems2015-03-30T03:31:28Z<p>Suli: /* Problem 6 */</p>
<hr />
<div>== Day 1 ==<br />
=== Problem 1 ===<br />
Let <math>p_1,p_2,p_3,...</math> be the prime numbers listed in increasing order, and let <math>x_0</math> be a real number between <math>0</math> and <math>1</math>. For positive integer <math>k</math>, define<br />
<br />
<math> x_{k}=\begin{cases}0&\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&\text{ if }x_{k-1}\ne0\end{cases} </math><br />
<br />
where <math>\{x\}</math> denotes the fractional part of <math>x</math>. (The fractional part of <math>x</math> is given by <math>x-\lfloor{x}\rfloor</math> where <math>\lfloor{x}\rfloor</math> is the greatest integer less than or equal to <math>x</math>.) Find, with proof, all <math>x_0</math> satisfying <math>0<x_0<1</math> for which the sequence <math>x_0,x_1,x_2,...</math> eventually becomes <math>0</math>.<br />
<br />
[[1997 USAMO Problems/Problem 1|Solution]]<br />
<br />
=== Problem 2 ===<br />
Let <math>ABC</math> be a triangle, and draw isosceles triangles <math>BCD, CAE, ABF</math> externally to <math>ABC</math>, with <math>BC, CA, AB</math> as their respective bases. Prove that the lines through <math>A,B,C</math> perpendicular to the lines <math>\overleftrightarrow{EF},\overleftrightarrow{FD},\overleftrightarrow{DE}</math>, respectively, are concurrent.<br />
<br />
[[1997 USAMO Problems/Problem 2|Solution]]<br />
<br />
=== Problem 3 ===<br />
Prove that for any integer <math>n</math>, there exists a unique polynomial <math>Q</math> with coefficients in <math>\{0,1,...,9\}</math> such that <math>Q(-2)=Q(-5)=n</math>.<br />
<br />
[[1997 USAMO Problems/Problem 3|Solution]]<br />
<br />
== Day 2 ==<br />
=== Problem 4 ===<br />
To ''clip'' a convex <math>n</math>-gon means to choose a pair of consecutive sides <math>AB, BC</math> and to replace them by three segments <math>AM, MN,</math> and <math>NC,</math> where <math>M</math> is the midpoint of <math>AB</math> and <math>N</math> is the midpoint of <math>BC</math>. In other words, one cuts off the triangle <math>MBN</math> to obtain a convex <math>(n+1)</math>-gon. A regular hexagon <math>P_6</math> of area <math>1</math> is clipped to obtain a heptagon <math>P_7</math>. Then <math>P_7</math> is clipped (in one of the seven possible ways) to obtain an octagon <math>P_8</math>, and so on. Prove that no matter how the clippings are done, the area of <math>P_n</math> is greater than <math>\frac{1}{3}</math>, for all <math>n\ge6</math>.<br />
<br />
[[1997 USAMO Problems/Problem 4|Solution]]<br />
<br />
=== Problem 5 ===<br />
Prove that, for all positive real numbers <math>a, b, c,</math><br />
<br />
<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>.<br />
<br />
[[1997 USAMO Problems/Problem 5|Solution]]<br />
<br />
=== Problem 6 ===<br />
Suppose the sequence of nonnegative integers <math>a_1,a_2,...,a_{1997}</math> satisfies <br />
<br />
<math>a_i+a_j \le a_{i+j} \le a_i+a_j+1</math><br />
<br />
for all <math>i, j \ge 1</math> with <math>i+j \le 1997</math>. Show that there exists a real number <math>x</math> such that <math>a_n=\lfloor{nx}\rfloor</math> (the greatest integer <math>\le x</math>) for all <math>1 \le n \le 1997</math>.<br />
<br />
[[1997 USAMO Problems/Problem 6|Solution]]<br />
<br />
== See Also ==<br />
{{USAMO newbox|year= 1997|before=[[1996 USAMO]]|after=[[1998 USAMO]]}}<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_7&diff=697292015 AIME II Problems/Problem 72015-03-29T22:49:17Z<p>Suli: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = w</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial<br />
<br />
Area(<math>PQRS</math>) = <math>\alpha w - \beta \cdot w^2</math>.<br />
<br />
Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
If <math>\omega = 25</math>, the area of rectangle <math>PQRS</math> is <math>0</math>, so<br />
<br />
<cmath>\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0</cmath><br />
<br />
and <math>\alpha = 25\beta</math>. If <math>\omega = \frac{25}{2}</math>, we can reflect <math>APQ</math> over PQ, <math>PBS</math> over <math>PS</math>, and <math>QCR</math> over <math>QR</math> to completely cover rectangle <math>PQRS</math>, so the area of <math>PQRS</math> is half the area of the triangle. Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>, <br />
<br />
<cmath> [ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90</cmath><br />
<br />
so<br />
<br />
<cmath>45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}</cmath><br />
<br />
and <br />
<br />
<cmath>\beta = \frac{180}{625} = \frac{36}{125}</cmath><br />
<br />
so the answer is <math>m + n = 36 + 125 = \boxed{161}</math>.<br />
<br />
==Solution 2==<br />
Similar triangles can also solve the problem.<br />
<br />
First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>BC</math> and solving. (The <math>8</math> side would be collinear with line <math>AB</math>)<br />
<br />
After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>. <br />
Solving for <math>BE</math> using the Pythagorean Formula, we get <math>BE = \frac{48}{5}</math>. We then know that <math>CE = \frac{77}{5}</math>.<br />
<br />
Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>. <br />
<br />
Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>.<br />
<br />
Let's work with <math>PF</math>. We know that <math>PQ</math> is parallel to <math>BC</math> so <math>\Delta APF</math> is similar to <math>\Delta ABE</math>. We can set up the proportion:<br />
<br />
<math>\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}</math>. Solving for <math>AF</math>, <math>AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega</math>. <br />
<br />
We can solve for <math>PS</math> then since we know that <math>PS=FE</math> and <math>FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega</math>.<br />
<br />
Therefore, <math>[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2</math>.<br />
<br />
This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>.<br />
<br />
- solution by abvenkgoo<br />
<br />
==Solution 3==<br />
Heron's Formula gives <math>[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90,</math> so the altitude from <math>A</math> to <math>BC</math> has length <math>\dfrac{2[ABC]}{BC} = \dfrac{36}{5}.</math><br />
<br />
Now, draw a parallel to <math>AB</math> from <math>Q</math>, intersecting <math>BC</math> at <math>T</math>. Then <math>BT = w</math> in parallelogram <math>QPBT</math>, and so <math>CT = 25 - w</math>. Clearly, <math>CQT</math> and <math>CAB</math> are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so<br />
<cmath>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.</cmath><br />
Solving gives <math>[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} = \dfrac{36w}{5} - \dfrac{36w^2}{125}</math>, so the answer is <math>36 + 125 = 161</math>.<br />
<br />
- solution by suli<br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_7&diff=697282015 AIME II Problems/Problem 72015-03-29T22:48:33Z<p>Suli: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = w</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial<br />
<br />
Area(<math>PQRS</math>) = <math>\alpha w - \beta \cdot w^2</math>.<br />
<br />
Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
If <math>\omega = 25</math>, the area of rectangle <math>PQRS</math> is <math>0</math>, so<br />
<br />
<cmath>\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0</cmath><br />
<br />
and <math>\alpha = 25\beta</math>. If <math>\omega = \frac{25}{2}</math>, we can reflect <math>APQ</math> over PQ, <math>PBS</math> over <math>PS</math>, and <math>QCR</math> over <math>QR</math> to completely cover rectangle <math>PQRS</math>, so the area of <math>PQRS</math> is half the area of the triangle. Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>, <br />
<br />
<cmath> [ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90</cmath><br />
<br />
so<br />
<br />
<cmath>45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}</cmath><br />
<br />
and <br />
<br />
<cmath>\beta = \frac{180}{625} = \frac{36}{125}</cmath><br />
<br />
so the answer is <math>m + n = 36 + 125 = \boxed{161}</math>.<br />
<br />
==Solution 2==<br />
Similar triangles can also solve the problem.<br />
<br />
First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>BC</math> and solving. (The <math>8</math> side would be collinear with line <math>AB</math>)<br />
<br />
After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>. <br />
Solving for <math>BE</math> using the Pythagorean Formula, we get <math>BE = \frac{48}{5}</math>. We then know that <math>CE = \frac{77}{5}</math>.<br />
<br />
Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>. <br />
<br />
Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>.<br />
<br />
Let's work with <math>PF</math>. We know that <math>PQ</math> is parallel to <math>BC</math> so <math>\Delta APF</math> is similar to <math>\Delta ABE</math>. We can set up the proportion:<br />
<br />
<math>\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}</math>. Solving for <math>AF</math>, <math>AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega</math>. <br />
<br />
We can solve for <math>PS</math> then since we know that <math>PS=FE</math> and <math>FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega</math>.<br />
<br />
Therefore, <math>[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2</math>.<br />
<br />
This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>.<br />
<br />
- solution by abvenkgoo<br />
<br />
==Solution 3==<br />
Heron's Formula gives <math>[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90,</math> so the altitude from <math>A</math> to <math>BC</math> has length <math>\dfrac{2[ABC]}{BC} = \dfrac{36}{5}.</math><br />
<br />
Now, draw a parallel to <math>AB</math> from <math>Q</math>, intersecting <math>BC</math> at <math>T</math>. Then <math>BT = w</math> in parallelogram <math>QPBT</math>, and so <math>CT = 25 - w</math>. Clearly, <math>CQT</math> and <math>CAB</math> are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so<br />
<math></math>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.<math><br />
Solving gives </math>[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} = \dfrac{36w}{5} - \dfrac{36w^2}{125}<math>, so the answer is </math>36 + 125 = 161$.<br />
<br />
- solution by suli<br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_8&diff=697272015 AIME II Problems/Problem 82015-03-29T22:18:47Z<p>Suli: /* Solution 2 (Proof without words) */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>a</math> and <math>b</math> be positive integers satisfying <math>\frac{ab+1}{a+b} < \frac{3}{2}</math>. The maximum possible value of <math>\frac{a^3b^3+1}{a^3+b^3}</math> is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution==<br />
Let us call the quantity <math>\frac{a^3b^3+1}{a^3+b^3}</math> as <math>N</math> for convenience. Knowing that <math>a</math> and <math>b</math> are positive integers, we can legitimately rearrange the given inequality so that <math>a</math> is by itself, which makes it easier to determine the pairs of <math>(a, b)</math> that work. Doing so, we have <cmath>\frac{ab+1}{a+b} < \frac{3}{2}</cmath> <cmath>\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2</cmath> <cmath>\implies a < \frac{3b - 2}{2b - 3}.</cmath> Now, observe that if <math>b = 1</math> we have that <math>N = \frac{a + 1}{a + 1} = 1</math>, regardless of the value of <math>a</math>. If <math>a = 1</math>, we have the same result: that <math>N = \frac{b + 1}{b + 1} = 1</math>, regardless of the value of <math>b</math>. Hence, we want to find pairs of positive integers <math>(a, b)</math> existing such that neither <math>a</math> nor <math>b</math> is equal to <math>1</math>, and that the conditions given in the problem are satisfied in order to check that the maximum value for <math>N</math> is not <math>1</math>.<br />
<br />
<br />
To avoid the possibility that <math>a = 1</math>, we want to find values of <math>b</math> such that <math>\frac{3b - 2}{2b - 3} > 2</math>. If we do this, we will have that <math>a < \frac{3b - 2}{2b - 3} = k</math>, where <math>k</math> is greater than <math>2</math>, and this allows us to choose values of <math>a</math> greater than <math>1</math>. Again, since <math>b</math> is a positive integer, and we want <math>b > 1</math>, we can legitimately multiply both sides of <math>\frac{3b - 2}{2b - 3} > 2</math> by <math>2b - 3</math> to get <math>3b - 2 > 4b - 6 \implies b < 4</math>. For <math>b = 3</math>, we have that <math>a < \frac{7}{3}</math>, so the only possibility for <math>a</math> greater than <math>1</math> is obviously <math>2</math>. Plugging these values into <math>N</math>, we have that <math>N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}</math>. For <math>b = 2</math>, we have that <math>a < \frac{4}{1} = 4</math>. Plugging <math>a = 2</math> and <math>b = 3</math> in for <math>N</math> yields the same result of <math>\frac{31}{5}</math>, but plugging <math>a = 2</math> and <math>b = 2</math> into <math>N</math> yields that <math>N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}</math>. Clearly, <math>\frac{31}{5}</math> is the largest value we can have for <math>N</math>, so our answer is <math>31 + 5 = \boxed{036}</math>.<br />
<br />
==Solution 2 (Proof without words)==<br />
<cmath>\frac{ab + 1}{a + b} < \frac{3}{2} \rightarrow 2ab + 2 < 3a + 3b,</cmath><br />
<cmath>\rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.</cmath><br />
<br />
<cmath>2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow</cmath><br />
<cmath>(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).</cmath><br />
<cmath>(a, b) = (2, 2), (2, 3), (3, 2).</cmath><br />
<cmath>\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.</cmath><br />
<cmath>\frac{31}{5} \rightarrow \boxed{36}.</cmath><br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_8&diff=697262015 AIME II Problems/Problem 82015-03-29T22:17:38Z<p>Suli: /* Solution 2 (Proof without words) */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>a</math> and <math>b</math> be positive integers satisfying <math>\frac{ab+1}{a+b} < \frac{3}{2}</math>. The maximum possible value of <math>\frac{a^3b^3+1}{a^3+b^3}</math> is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution==<br />
Let us call the quantity <math>\frac{a^3b^3+1}{a^3+b^3}</math> as <math>N</math> for convenience. Knowing that <math>a</math> and <math>b</math> are positive integers, we can legitimately rearrange the given inequality so that <math>a</math> is by itself, which makes it easier to determine the pairs of <math>(a, b)</math> that work. Doing so, we have <cmath>\frac{ab+1}{a+b} < \frac{3}{2}</cmath> <cmath>\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2</cmath> <cmath>\implies a < \frac{3b - 2}{2b - 3}.</cmath> Now, observe that if <math>b = 1</math> we have that <math>N = \frac{a + 1}{a + 1} = 1</math>, regardless of the value of <math>a</math>. If <math>a = 1</math>, we have the same result: that <math>N = \frac{b + 1}{b + 1} = 1</math>, regardless of the value of <math>b</math>. Hence, we want to find pairs of positive integers <math>(a, b)</math> existing such that neither <math>a</math> nor <math>b</math> is equal to <math>1</math>, and that the conditions given in the problem are satisfied in order to check that the maximum value for <math>N</math> is not <math>1</math>.<br />
<br />
<br />
To avoid the possibility that <math>a = 1</math>, we want to find values of <math>b</math> such that <math>\frac{3b - 2}{2b - 3} > 2</math>. If we do this, we will have that <math>a < \frac{3b - 2}{2b - 3} = k</math>, where <math>k</math> is greater than <math>2</math>, and this allows us to choose values of <math>a</math> greater than <math>1</math>. Again, since <math>b</math> is a positive integer, and we want <math>b > 1</math>, we can legitimately multiply both sides of <math>\frac{3b - 2}{2b - 3} > 2</math> by <math>2b - 3</math> to get <math>3b - 2 > 4b - 6 \implies b < 4</math>. For <math>b = 3</math>, we have that <math>a < \frac{7}{3}</math>, so the only possibility for <math>a</math> greater than <math>1</math> is obviously <math>2</math>. Plugging these values into <math>N</math>, we have that <math>N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}</math>. For <math>b = 2</math>, we have that <math>a < \frac{4}{1} = 4</math>. Plugging <math>a = 2</math> and <math>b = 3</math> in for <math>N</math> yields the same result of <math>\frac{31}{5}</math>, but plugging <math>a = 2</math> and <math>b = 2</math> into <math>N</math> yields that <math>N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}</math>. Clearly, <math>\frac{31}{5}</math> is the largest value we can have for <math>N</math>, so our answer is <math>31 + 5 = \boxed{036}</math>.<br />
<br />
==Solution 2 (Proof without words)==<br />
<cmath>\frac{ab + 1}{a + b} < \frac{3}{2} \rightarrow 2ab + 2 < 3a + 3b,</cmath><br />
<cmath>\rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.</cmath><br />
<br />
<math></math>2a - 3<math>, </math>2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow<cmath><br />
</cmath>(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).<cmath><br />
</cmath>(a, b) = (2, 2), (2, 3), (3, 2).<cmath><br />
</cmath>\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.<cmath><br />
</cmath>\frac{31}{5} \rightarrow \boxed{36}.<math></math><br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Sulihttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_8&diff=697252015 AIME II Problems/Problem 82015-03-29T22:16:32Z<p>Suli: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>a</math> and <math>b</math> be positive integers satisfying <math>\frac{ab+1}{a+b} < \frac{3}{2}</math>. The maximum possible value of <math>\frac{a^3b^3+1}{a^3+b^3}</math> is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution==<br />
Let us call the quantity <math>\frac{a^3b^3+1}{a^3+b^3}</math> as <math>N</math> for convenience. Knowing that <math>a</math> and <math>b</math> are positive integers, we can legitimately rearrange the given inequality so that <math>a</math> is by itself, which makes it easier to determine the pairs of <math>(a, b)</math> that work. Doing so, we have <cmath>\frac{ab+1}{a+b} < \frac{3}{2}</cmath> <cmath>\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2</cmath> <cmath>\implies a < \frac{3b - 2}{2b - 3}.</cmath> Now, observe that if <math>b = 1</math> we have that <math>N = \frac{a + 1}{a + 1} = 1</math>, regardless of the value of <math>a</math>. If <math>a = 1</math>, we have the same result: that <math>N = \frac{b + 1}{b + 1} = 1</math>, regardless of the value of <math>b</math>. Hence, we want to find pairs of positive integers <math>(a, b)</math> existing such that neither <math>a</math> nor <math>b</math> is equal to <math>1</math>, and that the conditions given in the problem are satisfied in order to check that the maximum value for <math>N</math> is not <math>1</math>.<br />
<br />
<br />
To avoid the possibility that <math>a = 1</math>, we want to find values of <math>b</math> such that <math>\frac{3b - 2}{2b - 3} > 2</math>. If we do this, we will have that <math>a < \frac{3b - 2}{2b - 3} = k</math>, where <math>k</math> is greater than <math>2</math>, and this allows us to choose values of <math>a</math> greater than <math>1</math>. Again, since <math>b</math> is a positive integer, and we want <math>b > 1</math>, we can legitimately multiply both sides of <math>\frac{3b - 2}{2b - 3} > 2</math> by <math>2b - 3</math> to get <math>3b - 2 > 4b - 6 \implies b < 4</math>. For <math>b = 3</math>, we have that <math>a < \frac{7}{3}</math>, so the only possibility for <math>a</math> greater than <math>1</math> is obviously <math>2</math>. Plugging these values into <math>N</math>, we have that <math>N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}</math>. For <math>b = 2</math>, we have that <math>a < \frac{4}{1} = 4</math>. Plugging <math>a = 2</math> and <math>b = 3</math> in for <math>N</math> yields the same result of <math>\frac{31}{5}</math>, but plugging <math>a = 2</math> and <math>b = 2</math> into <math>N</math> yields that <math>N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}</math>. Clearly, <math>\frac{31}{5}</math> is the largest value we can have for <math>N</math>, so our answer is <math>31 + 5 = \boxed{036}</math>.<br />
<br />
==Solution 2 (Proof without words)==<br />
<cmath>\frac{ab + 1}{a + b} < \frac{3}{2} \rightarrow 2ab + 2 < 3a + 3b,</cmath><br />
<cmath>\rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.</cmath><br />
<br />
<math>2a - 3</math>, <math>2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow</math><br />
<cmath>(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).</cmath><br />
<cmath>(a, b) = (2, 2), (2, 3), (3, 2).</cmath><br />
<cmath>\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{217}{35} = \frac{31}{5}.</cmath><br />
<cmath>\frac{31}{5} \rightarrow \boxed{36}.</cmath><br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Suli