Isosceles right triangle and tangency

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Isosceles right triangle and tangency

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Pan-American Girls’ Mathematical Olympiad, P2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

225 posts

jbaca

#1 h Oct 5, 2021, 9:24 PM • 1 Y

Y by HWenslawski

Consider the isosceles right triangle $ABC$ with $\angle BAC = 90^\circ$ . Let $\ell$ be the line passing through $B$ and the midpoint of side $AC$ . Let $\Gamma$ be the circumference with diameter $AB$ . The line $\ell$ and the circumference $\Gamma$ meet at point $P$ , different from $B$ . Show that the circumference passing through $A,\ C$ and $P$ is tangent to line $BC$ at $C$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

82 posts

#2 h Oct 5, 2021, 9:59 PM • 1 Y

Y by HWenslawski

Straightforward length bash work.
Let $AB = 2a$ , then $AM = MC = a$ . Note that $P$ is just the projection of $A$ wrt $BM$ . and we could easily get $BP = \frac{4a}{\sqrt{5}}$ and $PM = \frac{a}{\sqrt{5}}$ . Then, let $BP \cap (APC) = K$ . We could get by POP that
$PM \cdot MK = AM \cdot MC \implies MK = a \sqrt{5}$ Therefore, just note that
$BP \cdot BK = \frac{4a}{\sqrt{5}} \cdot 2a \sqrt{5} = 8a^2 = (2a \sqrt{2})^2 = BC^2$

Here's a synthetic solution.
Reflect $B$ wrt $M$ to get $K$ . Then, note that $\angle APB = 90^{\circ}$ , so $\angle APK = 90^{\circ}$ . But we know that $ABCK$ is a parallelogram since $MA = MC$ and $MB = MK$ . Therefore, $\angle ACK = 90^{\circ}$ . So, $APCK$ cyclic.
To finish, just note that
$BP \cdot BK = BP \cdot 2BM = 2 \cdot BA^2 = BC^2$ Note that $BP \cdot BM = BA^2$ is pop on $(APM)$ from point $B$ .

This post has been edited 1 time. Last edited by MathisWow, Oct 5, 2021, 10:01 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

future_gold_in_IMO

48 posts

future_gold_in_IMO

#3 h Oct 6, 2021, 2:24 AM

Y by

Consider $O$ the midpoint of $BC$ clearly $ABOP$ it's cyclic because $\angle BOA = 90^{\circ}$ = $\angle BPA$ , how $ABC$ it's isosceles $AO$ it's also the angle bissector of $\angle BAC$ , then $\angle BAO = 45^{\circ}$ = $\angle BPO$ , observe that $\angle MPO = 135^{\circ}$ and $\angle MCO = 45^{\circ}$ implies $OPMC$ cyclic.
Finally just note that $\angle OMC = 90^{\circ}$ = $\angle OPC$ give that $\angle PCO$ = $\angle POA$ = $\angle PBA$ and clearly $\angle PBA$ = $\angle PAC$ ( $APM$ and $ABM$ are similar) then $\angle PAC$ = $\angle PCO$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

49 posts

jampm

#4 h Oct 7, 2021, 12:59 AM

Y by

Angle chase Solution with nice construction

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

241 posts

AlanLG

#6 h Oct 7, 2021, 2:32 AM • 1 Y

Y by JuanDelPan

WLOG $AC=2$ , let $A(0,0)\Rightarrow B(0,2), C(2,0)$ , let $M(1,0)$ the midpoint of $AC$ and $P=(x,y)$ , the equation for the circle $\Gamma$ with center $(0,1)$ and radius 1 is

$(x-0)^2+(y-1)^2=1^2\Rightarrow x^2+(y-1)^2=1$
And the equation of $BM$ is

$y=\frac{2-0}{0-1}x+0\Rightarrow y=-2x+2$
Solving the sistem we have $x=\frac{4}{5}, y=\frac{2}{5}\Rightarrow P\left(\frac{4}{5}, \frac{2}{5}\right)$

If $O$ is the center of $(APB)$ , $O$ is of the form $O(1,k)$ (since $O$ is on the perpendicular bisector of AB)

Then

$AO=PO\Leftrightarrow (0-1)^2+(0-k)^2=(\frac{4}{5}-1)^2+(\frac{2}{5}-k)^2\Leftrightarrow k=1$
Finally, is enough to prove $BC^2=BO^2-r^2$ where $r$ is the radios of $(APB)$ , thus

$BC^2=BO^2-r^2\Leftrightarrow (0-2)^2+(2-0)^2=(0-1)^2+(2-(-1))^2-\left[(0-1)^2+(0-(-1))^2\right] \Leftrightarrow 8=8$
And we are done

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

38 posts

#7 h Oct 7, 2021, 2:39 AM

Y by

Notice that $AP'CP$ is a deltoid and not a parallelogram, and AP is perpendicular to $\ell$ .

jampm wrote:

Angle chase Solution with nice construction

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

11281 posts

#8 h Oct 7, 2021, 2:50 AM • 2 Y

Y by JuanDelPan, megarnie

Let $A=(0,0)$ , $B=(0,2)$ , $C=(2,0)$ in the Cartesian plane. We easily obtain $\ell:y=-2x+2$ and $\Gamma:x^2+(y-1)^2=1$ . Solving this system, we find $P=\left(\frac45,\frac35\right)$ . We can verify that $(x-1)^2+(y+1)^2=2$ is the circumcircle of $ACP$ , and $BC$ is $x+y=2$ . It suffices to show that the only solution to the system:
$\begin{align*}(x-1)^2+(y+1)^2&=2\\x+y&=2\end{align*}$ is $(x,y)=(2,0)$ . Substituting $y=2-x$ , we have $(x-1)^2+(x-3)^2=2$ , and so $2(x-2)^2=0$ . This finally proves the result.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

14 posts

#9 h Oct 7, 2021, 5:28 PM

Y by

Let $M$ be the midpoint of $AC$ , and let $D$ be the point such that $ABCD$ is a parallelogram, whose diagonals are clearly $\ell=BD$ and $AC$ . Since $AB$ is diameter in $\Gamma$ , we have $\angle APB=90^\circ=\angle APD$ , or $P$ is on circle $\Omega$ with diameter $AD$ . This circle clearly passes through $C$ because $ACD$ is by construction isoscelest at $C$ and with $\angle ACD=90^\circ$ . By symmetry in the isosceles right-angled triangle $ACD$ , $\Omega$ is tangent to the parallel to $AD$ through $C$ , which is $BC$ , and we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

302 posts

#10 h Oct 10, 2021, 5:01 AM

Y by

Why so many coord bash when you can angle chase?

It is sufficient to show $\angle PAC = \angle BCP$ . Let $K$ denote the midpoint of $BC$ . Notice that $\angle CMP = 90^{\circ} + \angle PAC$ and $\angle KMC = 90^{\circ}$ . So, it is sufficient to show $\angle PMK = \angle PCK \Leftrightarrow$ $CMPK$ cyclic. First, notice that $K$ lies on $(BPA)$ since $\angle BKA = 90^{\circ}$ . Now we have $45^{\circ} = \angle KCA = \angle BAK = \angle PAK$ as desired. $\blacksquare$

This post has been edited 2 times. Last edited by blackbluecar, Oct 10, 2021, 5:40 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

204 posts

#11 h Feb 25, 2022, 1:24 AM

Y by

Let $\angle PAC=\alpha ; \angle PCB=\beta$
Note that it is sufficient to show that: $\angle PAC=\angle PCB \iff \alpha=\beta$
$\implies \angle PAC=\angle PBA=\alpha \implies MP.MB=MA^2=MC^2 \implies 45-\alpha=\angle PBC=\angle PCA=45-\beta$
$\implies \alpha=\beta$ . $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6877 posts

#12 h Oct 26, 2022, 1:46 AM • 1 Y

Y by HamstPan38825

Here is one boring approach and one overkill approach.

Boring approach Set $\theta = \angle ABP = \arctan(1/2)$ , so $\angle PAB = 90^{\circ}-\theta$ and $\angle CAP = \theta$ . We wish to show $\angle BCP = \theta$ as well. By trigonometric Ceva, it is enough to verify $\frac{\sin(45^{\circ}-\theta)\sin(90^{\circ}-\theta)\sin(45^{\circ}-\theta)} {\sin\theta\sin\theta\sin\theta} = 1.$ Compute $\begin{align*} \sin\theta &= \frac{1}{\sqrt5} \\ \sin(90^{\circ}-\theta) &= \cos\theta = \frac{2}{\sqrt5} \\ \sin(45^{\circ}-\theta) &= \sin(45^{\circ})\cos\theta - \cos(45^{\circ})\sin\theta = \frac{1}{\sqrt2} \cdot \frac{2-1}{\sqrt5} = \frac{1}{\sqrt{10}} \end{align*}$ So the numerator and denominator both equal $5^{-3/2}$ as needed.

Overkill approach Define $Z$ as the second Brokard point --- in a general triangle, this point has barycentric coordinates $Z = \left( \frac{ab}{c} : \frac{bc}{a} : \frac{ca}{b} \right)$ thus always lying inside $ABC$ , and satisfies the angle condition $\angle ZBA = \angle ZAC = \angle ZCB.$ Now specialize to our right triangle to conclude that:

From $a=\sqrt2$ , $b=c=1$ we learn $Z = (2:1:2)$ , so $Z$ lies on the $B$ -median.
From $\angle ZBA = \angle ZAC$ follows $\angle BZA = \angle BAC = 90^{\circ}$ .

Hence $Z = P$ . Ergo, $\angle PAC = \angle PCB$ as needed.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a