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#1 h Dec 10, 2019, 1:22 AM • 5 Y

Y by noob_mathematician, kc5170, RedFireTruck, Adventure10, Rounak_iitr

In the triangle $\triangle ABC$ , let $G$ be the centroid, and let $I$ be the center of the inscribed circle. Let $\alpha$ and $\beta$ be the angles at the vertices $A$ and $B$ , respectively. Suppose that the segment $IG$ is parallel to $AB$ and that $\beta = 2\tan^{-1}(1/3)$ . Find $\alpha$ .

This post has been edited 1 time. Last edited by djmathman, Sep 14, 2020, 1:16 PM
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awesomemathlete

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awesomemathlete

#2 h Dec 10, 2019, 1:22 AM • 4 Y

Y by noob_mathematician, Agsh2005, RedFireTruck, Adventure10

By the angle bisector theorem and mass points, $IG$ being parallel with $AB$ is equivalent with $c=\frac{a+b}{2}$ and also $3r=h_c$ but then $\tan{\frac{\beta}{2}}=\frac{1}{3}=\frac{r}{s-b}$ implies $3r=h_c=s-b=\frac{3a-b}{4}$ because the lengths of the tangents from the vertices to the incircle are $s-a$ , $s-b$ , and $s-c$ from $A$ , $B$ , and $C$ respectively. This implies that:

$S=\frac{ch_c}{2}=\frac{(a+b)(3a-b)}{16}=\sqrt{\frac{3a+3b}{4}\cdot\frac{3a-b}{4}\cdot\frac{3b-a}{4}\cdot\frac{a+b}{4}}=\sqrt{s(s-a)(s-b)(s-c)}\implies$
$(a+b)(3a-b)=\sqrt{(3a+3b)(3a-b)(3b-a)(a+b)}=(a+b)\sqrt{3(3a-b)(3b-a)}\implies$
$3a-b=\sqrt{3(3a-b)(3b-a)}\implies$
$\sqrt{3a-b}=\sqrt{3(3b-a)}\implies$
$3a-b=3(3b-a)=9b-3a\implies$
$3a=5b$

So $ABC$ is a $3-4-5$ triangle and $\alpha=\boxed{\frac{\pi}{2}}$ .
$\square$

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Jorvis

#4 h Dec 10, 2019, 1:32 AM • 3 Y

Y by RedFireTruck, Adventure10, Mango247

Cartesian coordinates also work, with surprisingly little computation.

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jeff10

#5 h Dec 10, 2019, 1:43 AM • 4 Y

Y by noob_mathematician, RedFireTruck, Adventure10, Mango247

Sketch

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#6 h Dec 10, 2019, 2:00 AM • 3 Y

Y by noob_mathematician, RedFireTruck, Adventure10

Length bashing for the win!

Denote: $E$ as perpendicular from $I$ to $AB$ , $F$ as $IE$ intersect $BC$ , and let line passing through $C$ and passing through $AB$ meet $IE$ at $H$ , so $\angle AHE=90^{\circ}$ . Also let $AG$ meet $AB$ at $M$ , $AI$ meet $AB$ at $D$ , and let $IE=1$ . Then:

1. $EB=3$ .
2. $\tan(\beta)=\frac 34$ , so $FE=\frac 94$ .
3. $\frac{AG}{AM}=\frac{AI}{AD}=\frac{HI}{HE}=\frac 23$ (the well-known centroid property).
4. This means $HI=2, HE=3$ . This means $HF=\frac 34$ .
5. $AHE$ and $BEF$ are similar so $AH=1$ .
6. Now $\tan AIH=\frac 12=\tan IED$ .
7. But then $\tan BID=\tan (BIE - EID) = (3-\frac 12)/(1+3\cdot \frac 12)=1$ so $BID=45^{\circ}$ .
8. $\alpha/2=90^{\circ}-\angle BID$ so $\alpha=90^{\circ}$ , i.e. $\frac{\pi}{2}$ .

This post has been edited 1 time. Last edited by Anzoteh, Dec 10, 2019, 2:21 AM
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#7 h Dec 10, 2019, 2:52 AM • 4 Y

Y by noob_mathematician, RedFireTruck, Adventure10, Mango247

If $CC’$ is the bisectrix of the angle $ACB$ and $I$ is the center of the inscribed circle, then $\frac{CI}{IC’} = \frac{CA + CB}{AB}$ , hence $IG \| AB$ if and only if $CA + CB = 2AB$ . Fix $A$ and $B$ ; then $C$ is on an ellipse with foci $A$ and $B$ . Since the angle $CBA$ is determined, with $cos\beta = 4/5$ , $C$ is also on a half line starting at $B$ . That half line intersects the ellipse in a unique point - hence the triangle $ABC$ is determined up to similarities. A right triangle with $AB = 4$ , $AC = 3$ , and $BC = 5$ works; therefore $\alpha = \pi/2$ .

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#8 h Dec 10, 2019, 4:33 AM • 3 Y

Y by noob_mathematician, RedFireTruck, Adventure10

The condition $IG\parallel AB$ is equivalent to $\frac{r}{h_C}=\frac{1}{3}$ , which (using $K=rs=\frac{ch_C}{2}$ ) implies that $a+b=2c$ . So $\sin\alpha+\sin\beta=2\sin\gamma$ where $\gamma=\angle{C}$ .

We apply Vincent Huang Bashing. Let $x=e^{i\alpha},y=e^{i\beta},z=e^{i\gamma}$ so that $xyz=-1$ . Then
$\frac{x-x^{-1}}{2i}+\frac{y-y^{-1}}{2i}=\frac{z-z^{-1}}{i}$ so
$(2y-1)x-(y-y^{-1})+(1-2y^{-1})x^{-1}=0.$ With $y=\frac{(3+i)^2}{10}=\frac{4}{5}+i\frac{3}{5}$ , this gives
$\left(x-i\right)\left(\left(\frac{3}{5}+i\frac{6}{5}\right)-\left(\frac{6}{5}+i\frac{3}{5}\right)\right)=\left(\frac{3}{5}+i\frac{6}{5}\right)x-i\frac{6}{5}+\left(-\frac{3}{5}+i\frac{6}{5}\right)x^{-1}=0$ so $x=i$ and thus $\alpha=\boxed{\frac{\pi}{2}}$ .

To learn more about this amazing method, see this handout! :love:

:love:

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yayups

#9 h Dec 10, 2019, 5:10 AM • 4 Y

Y by noob_mathematician, RedFireTruck, Adventure10, Mango247

Claim: In a triangle $ABC$ , we have $IG\parallel AB$ if and only if $CA+CB=2\cdot AB$ .

Proof: Note that $IG\parallel AB$ if and only if $[IAB]=[GAB]$ . We see that $[IAB]:[IAC]:[IBC]=AB:AC:BC$ , so $[IAB]=\frac{AB}{AB+AC+BC}[ABC].$ But it's also well known that $[GAB]=\frac{1}{3}[ABC]$ , so $[IAB]=[GAB]\iff\frac{AB}{AB+AC+BC}[ABC]=\frac{1}{3}[ABC]\iff CA+CB=2\cdot AB,$ as desired. $\blacksquare$

The real surprise comes in the answer extraction step. Note that $\tan\beta=\frac{2\cdot(1/3)}{1-(1/3)^2}=\frac{3}{4},$ so $\beta$ is the smallest angle of a $3$ - $4$ - $5$ triangle.

Remark: At this point, one may notice that the $3$ - $4$ - $5$ triangle satisfies the condition of the lemma, so the answer is just $\alpha=\pi/2$ . Proving that this value is a monotonicity argument that needs a bit of care, but isn't too difficult. However, we proceed with a mindless bash that doesn't require this insight.

By the law of sines, the condition is equivalent to $\sin\alpha+\sin\beta=2\sin(\alpha+\beta)$ . Expanding both sides and using the values of $\sin\beta$ and $\cos\beta$ , we get $\sin\alpha+\frac{3}{5}=\frac{8}{5}\sin\alpha+\frac{6}{5}\cos\alpha,$ or $2\cos\alpha=1-\sin\alpha$ . Squaring both sides, we get $4(1-\sin^2\alpha)=1-2\sin\alpha+\sin^2\alpha,$ or $5\sin^2\alpha-2\sin\alpha-3=0$ . This gives $\alpha=\frac{2\pm\sqrt{4+4\cdot 3\cdot 5}}{10}=\frac{2\pm 8}{10}=1,-3/5.$ Since $\alpha\in(0,\pi)$ , we must have $\sin\alpha=1$ , so $\boxed{\alpha=\pi/2}$ .

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#10 h Dec 10, 2019, 6:15 PM • 3 Y

Y by RedFireTruck, Adventure10, Levieee

i coordinate bashed lol.

So let $B=(0,0)$ and $C= \left(x, \dfrac{3}{4} x\right)$ and $A = (\gamma x,0)$ . we used the fact $\tan(\beta) = \dfrac{3}{4}$ . clearly $G=(\dfrac{x}{3}, \dfrac{x}{4})$ and since $GI$ is parallel $AB$ , we must have the $y$ coordinate of $I$ is $\dfrac{x}{4}$ . this means $r=\dfrac{x}{4}$ . by using $rs=[ABC]$ , $\dfrac{x}{4}\left(\dfrac{\dfrac{5}{4}x+\gamma x +x \sqrt{(1-\gamma)^2+\dfrac{9}{16}}}{2}\right) = \dfrac{1}{2} \gamma x \dfrac{3}{4} x \to \gamma = 1$ . hence the triangle is obviously a $3-4-5$ triangle, making $\alpha=\dfrac{\pi}{2}$

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#11 h Dec 10, 2019, 7:07 PM • 4 Y

Y by noob_mathematician, RedFireTruck, Adventure10, Mango247

First, let us determine the value of $\tan \beta$ . By the double-angle formula for tangent, we have

$\begin{align*} \tan \beta &=\tan\left(2\tan^{-1}(1/3)\right) \\ &=\dfrac{2\left(\dfrac{1}{3}\right)}{1-\left(\dfrac{1}{3}\right)^2} \\ &=\dfrac{2/3}{8/9} \\ &=\dfrac{3}{4}. \end{align*}$
We will next prove that the condition $\overline{IG} \parallel \overline{AB}$ necessarily implies $AC+BC=2AB$ .

Draw the angle bisector and median from $C$ and let them hit $AB$ at points $D$ and $M$ , respectively. These segments, of course, pass through $I$ and $G$ , respectively, and since $\overline{IG} \parallel \overline{AB}$ , we find that $\triangle CIG \sim \triangle CDM$ . Hence, $\dfrac{CI}{ID}=\dfrac{CG}{GM}=2$ by the centroid rule.

Now, we apply a common technique used in Euclidean geometry when dealing with angle bisectors and draw $AI$ and $BI$ . Since these are also angle bisectors, we can invoke the Angle Bisector Theorem on triangles $ACD$ and $BCD$ to get $\dfrac{AC}{AD}=\dfrac{CI}{ID}=\dfrac{BC}{BD}=2.$ Therefore, $AC+BC=2(AD+BD)=2AB$ , as claimed.

The next step is to draw the altitude $CE$ . Since $\tan \beta=\dfrac{3}{4}$ , we can let $CE=3x$ , $BE=4x$ , and $BC=5x$ for some real number $x$ , as $\triangle BCE$ is a $3-4-5$ right triangle. Furthermore, let $AE=y$ so that $AC=\sqrt{9x^2+y^2}$ .

Using the observation that $AC+BC=2AB$ gives

$\begin{align*} \sqrt{9x^2+y^2}+5x &=2\left(4x+y\right) \\ \Rightarrow \sqrt{9x^2+y^2} &=8x+2y \\ \Rightarrow \sqrt{9x^2+y^2} &=3x+2y \\ \Rightarrow 9x^2+y^2 &=\left(3x+2y\right)^2 \\ \Rightarrow 9x^2+y^2 &=9x^2+12xy+4y^2 \\ \Rightarrow 12xy+3y^2 &=0 \\ \Rightarrow y\left(4x+y\right) &=0. \end{align*}$
Thus, either $y=0$ or $4x+y=0$ . But the latter is impossible, since $4x+y=AB$ and $AB=0$ would result in a degenerate triangle. We conclude, then, that $y=0$ , so the altitude from $C$ actually hits $AB$ at $A$ , and $ABC$ is, in fact, a $3-4-5$ right triangle. Therefore, $\angle A=90^\circ=\boxed{\dfrac{\pi}{2}}$ .

This post has been edited 1 time. Last edited by CountofMC, Dec 10, 2019, 7:18 PM

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ayan_mathematics_king

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ayan_mathematics_king

#12 h Jan 30, 2020, 8:57 AM • 4 Y

Y by noob_mathematician, RedFireTruck, Adventure10, Mango247

Slightly different from the other solutions and may be more elegant

$D$ is the midpoint of $BC$
Extend $IG$ to meet $BC$ at $Q$ . Clearly, $\frac{BQ}{QD} =2$ .
So, $BQ=\frac{2}{3}\cdot \frac{a}{2}=\frac{a}{3}$ .
By simple trigonometry we have $\cos{B}=\frac{4}{5}$ , $\sin{B}=\frac 35$ , $\tan{B/2}=\frac 13$ ,
$\sin{B/2}=\frac{1}{\sqrt{10}}$
Notice that $\angle IBQ=\angle BIQ$ .
So by cosine rule in $BIQ$ we get,
$BI^2=2BQ^2(1+\cos{B})=\frac{2a^2}{5}$ .
But we have $BI^2=\frac{r^2}{\sin^2{B/2}}=10r^2$ .
Comparing the value of $BI$ we get $r=\frac{a}{5}$ .
Which implies, $\frac{\Delta}{s}=\frac{a}{5} \implies \frac{ac\sin{B}}{a+b+c}=\frac{a}{5}$ .
Simplifying this we have , $a+b=2c$ .
Without loss of generality let $c=1$ . $a+b=2$ .
By cosine rule on $\Delta ABC$ we have $a^2+1-b^2=\frac{8}{5}\cdot a$ .

Solving these two equations we get $b=\frac 34$ and $a=\frac 54$ . Clearly it satisfies Pythagoras theorem. Hence , $\alpha=\frac{\pi}{2}$ $\blacksquare$

This post has been edited 3 times. Last edited by ayan_mathematics_king, Dec 31, 2020, 1:16 PM

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#13 h Apr 12, 2020, 1:04 PM • 1 Y

Y by RedFireTruck

From Heron's formula we have $r^2s^2=s(s-a)(s-b)(s-c)$ . Since $IG \parallel AB \implies A_{ABI}=A_{ABG}=\frac{A_{ABC}}{3}=\frac{rs}{3}$
From this we get that $c=\frac{a+b}{2}, s-c=\frac{a+b}{4}$ and $3r^2=(s-a)(s-b)$
Incircle intersects line $AB$ in point $D$ , so $DA=s-a=\frac{r}{\tan \frac{\alpha}{2}}$ and $DB=s-b=\frac{r}{\tan \frac{\beta}{2}}$
So, from factorization $\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}=\frac{1}{3}$ . Also, $\tan \frac{\beta}{2}=\frac{1}{3}$ . So:
$\tan \frac{\alpha}{2}=1 \implies \frac{\alpha}{2}=\frac{\pi}{4}$ , and finally $\alpha=\frac{\pi}{2}$ . $Q.E.D.$

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#14 h Jun 21, 2020, 7:20 PM • 2 Y

Y by RedFireTruck, tapir1729

Could I possibly have some feedback on the following solution? I'm unfamiliar with the Putnam scoring process (although I know it's very strict). Does this solution work? Are there any stylistic improvements that could be made? Thanks!

$[asy] pair A = (0,0), B = (4,0), C = (4,3); draw(A--B--C--cycle); pair G = centroid(A,B,C), I = incenter(A,B,C); draw(A--I--(3,0),red+dashed); draw(incircle(A,B,C)); draw(G--(4,1),red); dot(A^^B^^C); dot(G^^I^^(4,1),red); label("$A$", B, dir(-45)); label("$B$", A, dir(-140)); label("$C$", C, dir(60)); label("$G$", G, dir(110)); label("$I$", I, dir(80)); label("$P$", (4,1), dir(0)); [/asy]$

Scale so $AB=4$ . Place the triangle in the coordinate plane such that $B=(0,0)$ and $A=(4,0)$ . Note that $2\tan^{-1}(1/3)=\tan^{-1}(3/4)$ , so $C$ lies on the line $y=\tfrac{3}{4}x$ . Let $C=(4a,3a)$ for some $a$ . Furthermore, $I$ lies on the bisector of $\angle B$ , which has equation $y=\tfrac{1}{3}x$ . Let $I=(3b,b)$ for some $b$ .

Since $\overline{AB}$ is now horizontal, the condition is equivalent to $I$ and $G$ having the same $y$ -coordinate. It is well-known that the coordinates of the centroid of a triangle is equal to the average of the coordinates of its vertices, so $G=(\tfrac{4+4a}{3},a)$ , from which we get $a=b$ by comparing with $I$ .

Now, note that the inradius of $\triangle ABC$ is the distance from $I$ to $\overline{AB}$ , which is $a$ . Consider the point $P=(4a,a)$ , which lies on the incircle of $\triangle ABC$ . Note that $\overline{IP}$ is horizontal while $\overline{CP}$ is vertical, so $\angle IPC$ is right and $\overline{CP}$ is tangent to the incircle of $\triangle{ABC}$ , which implies $P$ lies on $\overline{AC}$ , so $\overline{AC}$ is vertical and $\alpha=\boxed{90^{\circ}}$

This post has been edited 1 time. Last edited by Archeon, Jun 22, 2020, 12:22 PM

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#15 h Dec 1, 2020, 3:46 AM • 4 Y

Y by RedFireTruck, Mango247, Mango247, Mango247

Archeon wrote:

Could I possibly have some feedback on the following solution? I'm unfamiliar with the Putnam scoring process (although I know it's very strict). Does this solution work? Are there any stylistic improvements that could be made? Thanks!

To be honest, I very much favor your solution to the others, simply because it's easier to understand. But I don't know how the Putnam graders would like a solution that uses specific values off the bat. I have just noticed that solutions to Putnam geometry problems always remain pretty general and arbitrary. Maybe I'm wrong and your proof is perfect as it is, but good job!

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#16 h Jan 28, 2021, 11:04 AM • 1 Y

Y by RedFireTruck

Lemma 1: Let MN be the line parallel to AB through I where M lies on AC and N lies on BC . Then $MN= \frac{c(a+b)}{a+b+c}$
Proof: It is not difficult to see that $MN= c- r cot(A) - r cot(B)$
We now use trig identities to get $MN=c- \frac{(s-a)^2- r^2}{2(s-a)} - \frac{(s-b)^2-r^2}{2(s-b)}= \frac{c -((s-a)(s-b)*c-cr^2)}{2(s-a)(s-b)}$
$= c -(\frac{c}{2} - \frac{c(s-c)}{2s}) = \frac{c}{2}+ \frac{c(s-c)}{2s} = \frac{c(a+b)}{a+b+c}$

Now that we have proven the above lemma we will now use the condition that $IG \vert \vert AB$ . Now a line passing through G divides AC and BC in the ratio $2:1$ . By similarity we say that $MN:AB= 2:3 \implies \frac{a+b}{a+b+c}= \frac{2}{3}\implies c=\frac{a+b}{2}$

Now the ebove relation on lengths reduces to a relation on angles, that is $sin(C) = \frac{sin(\alpha)+sin(\beta)}{2}$
$sin(\pi - \alpha - \beta) =\frac{sin(\alpha) +sin(\beta)}{2}= sin(\alpha + \beta)\implies sin(\alpha) + \frac{3}{5} = 2(sin(\alpha)\frac{4}{5} + cos(\alpha)\frac{3}{5}) \implies cos(\alpha)+1 = sin(\alpha)$ Now the only solution of the last equation(given that $\alpha,\beta$ are angles of a trinage) is $\alpha = \frac{\pi}{2}$ and we are done!

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#17 h Apr 13, 2024, 12:47 PM

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Note that $\cos \beta = \frac{4}{5}$ . As $IG||AB$ and $G$ divides the median in ratio $2:1$ , we get that $a+b=2c$ . Now by Cosine Rule,
$\frac{4}{5}=\cos \beta = \frac{a^2+c^2-b^2}{2ac} \stackrel{\text{magic}}{\implies} \frac{b}{a} = \frac{3}{5} \implies \frac{AC}{BC} = \sin \beta$ This directly implies that $ABC$ is right angled at $A$ i.e. $\alpha = 90^o$ . $\blacksquare$

This post has been edited 1 time. Last edited by sanyalarnab, Apr 13, 2024, 12:47 PM

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#18 h Apr 13, 2024, 5:53 PM

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Since $IG||AB$ , the $C-$ altitude is equivalent to three times the inradius. Thus, we have $3r\cdot AB=(AB+BC+AC)r, AC+BC=2AB$

Let $\tan \angle{\frac{CAB}{2}}=\frac{1}{x}, \tan \angle{\frac{ACB}{2}}=\frac{1}{y}$

We have $x+2y+3=2x+6, x+3=2y$

By tangent addition formula, we have $\frac{x+y}{xy-1}=\tan{(\frac{\pi}{2}-\arctan{\frac{1}{3}})}=3$

Plug $x+3=2y$ into the system to solve $y=2, x=1\implies \tan \angle{\frac{CAB}{2}}=\frac{1}{x}=1, \angle{CAB}=\alpha=90^{\circ}$

This post has been edited 1 time. Last edited by Bluesoul, Apr 13, 2024, 9:38 PM

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lifeismathematics

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lifeismathematics

#19 h May 8, 2024, 6:30 AM

Y by

Denote the projection of $C$ onto $AB$ as $D$ , and suppose projection of $I$ onto $AB$ be $E$ , $IG \cap CB:=\{F\}, CG \cap{AB}:=\{M\} , CD \cap IG:=\{T\}$ , then since $IG \parallel AB$ we have $\triangle{CGF} \sim \triangle{CMB}$ and hence we have $\frac{CD}{CT}=\frac{CM}{CG}=\frac{2}{3}$ , now since $DT \parallel IE$ and $IG \parallel AB$ we must have $IE=DT=r$ (inradius of the triangle). SO we have $CD=3r$ , now In $\triangle{CDB}$ we have $\frac{CD}{DB}=\tan(\beta) \implies \frac{3r}{DB}=\frac{2\cdot \frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}=\frac{3}{4} \implies DB=4r$ , now in $\triangle{IBE}$ we have $\frac{IE}{EB}=\tan\left(\frac{\beta}{2}\right)$ , hence we have $EB=3r$ , now that implies $DE=DB-EB=r$ , now that says that $TI \perp IE$ which gives the incircle of $\triangle{ABC}$ is tangent to $CD$ , which in tur implies $D$ lies on $AC$ and hence $\alpha=90^{\circ}$ . $\square$

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#20 h Apr 27, 2025, 7:55 PM

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We claim the answer is $\boxed{\alpha=90^{\circ}}.$
$[asy] pair B=(0,0), A=(4,0), C=(4,3), I=(3,1), F=(5/2,0), D=(4/3, 1), E=(4,1); draw(A--B--C--cycle); draw(D--E); draw(C--F); draw(B--I--A); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, dir(0)); label("$I$", I+(0.1,0), NE); label("$B$", B, SW); label("$F$", F, S); label("$A$", A, SE); pair G=(A+B+C)/3, M=(2,0); draw(C--M); label("$G$", G, NW); label("$M$", M, S); [/asy]$
Let $IG$ intersect $BC$ and $AC$ at $D$ and $E,$ respectively, let $CI$ and $CG$ intersect $AB$ at $F$ and $M,$ respectively, and let $BC=a$ and $AC=b.$ Observe that since $CM$ is a median of $\triangle ABC$ and $G$ is the centroid, we have $\dfrac{CG}{CM}=\dfrac{2}{3}.$ Then, since $DG\parallel BM,$ $\triangle CDG\sim \triangle CBM,$ so $\dfrac{CD}{CB}=\dfrac{CD}{a}=\dfrac{CG}{CM}=\dfrac{2}{3}.$ Similarly, since $GE\parallel FA,$ we have $\triangle CGE\sim \triangle CMA,$ so $\dfrac{CE}{CA}=\dfrac{CE}{b}=\dfrac{CG}{CM}=\dfrac{2}{3}.$
$[asy] pair B=(0,0), A=(4,0), C=(4,3), I=(3,1), F=(5/2,0), D=(4/3, 1), E=(4,1); draw(A--B--C--cycle); draw(D--E); draw(C--F); draw(B--I--A); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, dir(0)); label("$I$", I, NW); label("$B$", B, SW); label("$F$", F, S); label("$A$", A, SE); [/asy]$
Now, since $DI\parallel BF,$ $\triangle CDI\sim \triangle CBF,$ so we have $\dfrac{CI}{CF}=\dfrac{CD}{CB}=\dfrac{2}{3},$ so $CI=\dfrac{2CF}{3}$ and $IF=CF-CI=\dfrac{CF}{3}.$

Since $I$ is the incenter of $\triangle ABC,$ $BI$ bisects $\angle ABC$ and $AI$ bisects $\angle BAC,$ so by the angle bisector theorem, $\dfrac{BF}{a}=\dfrac{BF}{BC}=\dfrac{IF}{CI}=\dfrac{\left(\frac{CF}{3}\right)}{\left(\frac{2CF}{3}\right)}=\dfrac{1}{2}=\dfrac{FA}{AC}=\dfrac{FA}{b},$ so $BF=\dfrac{a}{2}$ and $FA=\dfrac{b}{2}.$ As such, $AB=BF+FA=\dfrac{a}{2}+\dfrac{b}{2}.$

By the double angle formula, $\dfrac{\sin(\beta)}{\cos(\beta)}=\tan(\beta)=\dfrac{\frac{1}{3}+\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}=\dfrac{\left(\frac{2}{3}\right)}{\left(\frac{8}{9}\right)}=\dfrac{3}{4}.$ As such, $\sin(\beta)=\dfrac{3\cos(\beta)}{4},$ so $\sin^2(\beta)+\cos^2(\beta)=\dfrac{9\cos^2(\beta)}{16}+\cos^2(\beta)=\dfrac{25\cos^2(\beta)}{16}=1,$ so $\cos^2(\beta)=\dfrac{16}{25}$ so $\cos(\beta)=\pm\dfrac{4}{5}.$ If $\cos(\beta)=-\dfrac{4}{5},$ then $\sin(\beta)=\dfrac{3\cos(\beta)}{4}=-\dfrac{3}{5},$ but since $0\leq \beta \leq 180^{\circ},$ $\sin(\beta)>0,$ so $\cos(\beta)\neq -\dfrac{4}{5}.$ As such, $\cos(\beta)=\dfrac{4}{5},$ and $\sin(\beta)=\dfrac{3\cos(\beta)}{4}=\dfrac{3}{5}.$

Now, by the law of cosines, $b^2=AC^2=BC^2+AB^2-2\cdot AB\cdot BC\cos(\beta)=a^2+\left(\dfrac{a}{2}+\dfrac{b}{2}\right)^2-2\cdot \left(\dfrac{a}{2}+\dfrac{b}{2}\right)\cdot a\cdot \dfrac{4}{5}=a^2+\dfrac{a^2}{4}+\dfrac{ab}{2}+\dfrac{b^2}{4}-\dfrac{4a^2}{5}-\dfrac{4ab}{5}=\dfrac{9a^2}{20}-\dfrac{3ab}{10}+\dfrac{b^2}{4},$ so $\dfrac{3b^2}{4}+\dfrac{3ab}{10}-\dfrac{9a^2}{20}=0,$ so $5b^2+2ab-3a^2=(5b-3a)(b+a)=0.$ Since $a,b>0,$ $a+b\neq 0,$ so $5b=3a.$

As such, by the law of sines, $\dfrac{a}{\sin(\alpha)}=\dfrac{b}{\sin(\beta)},$ so $\dfrac{3}{5}=\dfrac{b}{a}=\dfrac{\sin(\beta)}{\sin(\alpha)}=\dfrac{3}{5\sin(\alpha)},$ so $\sin(\alpha)=1.$ Since $0\leq \alpha\leq 180^{\circ},$ we must have $\alpha=\boxed{90^{\circ}},$ as desired. $\Box$

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